Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "2020~2021学年9月湖北宜昌伍家岗区宜昌英杰学校初一上学期月考（英杰教育集团）第11题3分", "初一上学期单元测试《解读绝对值》第3题", "2017年湖南长沙天心区初二竞赛长郡教育集团抵达杯第1题5分", "2019~2020学年河南郑州金水区郑州市第八中学初一上学期期中第8题3分" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$或$$-1$$ " } ], [ { "aoVal": "C", "content": "$$2$$或$$-2$$ " } ], [ { "aoVal": "D", "content": "$$0$$或$$-2$$ " } ] ]

[ "课内体系->知识点->数->有理数->绝对值->绝对值综合", "课内体系->能力->推理论证能力" ]

[ "∵$$a+b+c=0$$且$$a$$、$$b$$、$$c$$均$$\\ne 0$$， ∴$$a$$、$$b$$、$$c$$三数符号为两正一负或两负一正， 不妨设， ①$$a$$，$$b$$为正，$$c$$为负，此时$$abc$$为负： $$\\frac{a}{\\left\\textbar{} a \\right\\textbar}+\\frac{b}{\\left\\textbar{} b \\right\\textbar}+\\frac{c}{\\left\\textbar{} c \\right\\textbar}+\\frac{abc}{\\left\\textbar{} abc \\right\\textbar}$$ $$=1+1+(-1)+(-1)$$ $$=0$$． ②$$a$$，$$b$$为负，$$c$$为正，此时$$abc$$为正， $$\\frac{a}{\\left\\textbar{} a \\right\\textbar}+\\frac{b}{\\left\\textbar{} b \\right\\textbar}+\\frac{c}{\\left\\textbar{} c \\right\\textbar}+\\frac{abc}{\\left\\textbar{} abc \\right\\textbar}$$ $$=-1+(-1)+1+1$$ $$=0$$． 综上原式$$=0$$，故选$$\\text{A}$$． " ]

[ "2002年第19届全国初中数学联赛竞赛第6题7分" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "竞赛->知识点->数论->同余->完全平方数" ]

[ "$$3n+1$$是一个完全平方数，那这个完全平方数$$a$$显然不是$$3$$的倍数． 设$$a=3t\\pm 1$$，则$${{a}^{2}}=9{{t}^{2}}\\pm 6t+1$$， $${{a}^{2}}=3n+1=9{{t}^{2}}\\pm 6t+1$$，$$\\textasciitilde\\Rightarrow n=3{{t}^{2}}\\pm 2t$$， $$n+1=3{{t}^{2}}\\pm 2t+1={{(t\\pm 1)}^{2}}+{{t}^{2}}+{{t}^{2}}$$． 而若$$k$$为$$1$$，当$$n=21$$时$$3n+1=64$$是完全平方数，但是$$n+1=22$$不能写成一个或两个完全平方数的和，故$$k$$最小为$$3$$，选择$$\\text{C}$$． " ]

[ "初一上学期单元测试《解读绝对值》第3题", "2017年湖南长沙天心区初二竞赛长郡教育集团抵达杯第1题5分", "2020~2021学年9月湖北宜昌伍家岗区宜昌英杰学校初一上学期月考（英杰教育集团）第11题3分", "2019~2020学年河南郑州金水区郑州市第八中学初一上学期期中第8题3分" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$或$$-1$$ " } ], [ { "aoVal": "C", "content": "$$2$$或$$-2$$ " } ], [ { "aoVal": "D", "content": "$$0$$或$$-2$$ " } ] ]

[ "课内体系->能力->推理论证能力", "课内体系->知识点->数->有理数->绝对值->绝对值综合" ]

[ "∵$$a+b+c=0$$且$$a$$、$$b$$、$$c$$均$$\\ne 0$$， ∴$$a$$、$$b$$、$$c$$三数符号为两正一负或两负一正， 不妨设， ①$$a$$，$$b$$为正，$$c$$为负，此时$$abc$$为负： $$\\frac{a}{\\left\\textbar{} a \\right\\textbar}+\\frac{b}{\\left\\textbar{} b \\right\\textbar}+\\frac{c}{\\left\\textbar{} c \\right\\textbar}+\\frac{abc}{\\left\\textbar{} abc \\right\\textbar}$$ $$=1+1+(-1)+(-1)$$ $$=0$$． ②$$a$$，$$b$$为负，$$c$$为正，此时$$abc$$为正， $$\\frac{a}{\\left\\textbar{} a \\right\\textbar}+\\frac{b}{\\left\\textbar{} b \\right\\textbar}+\\frac{c}{\\left\\textbar{} c \\right\\textbar}+\\frac{abc}{\\left\\textbar{} abc \\right\\textbar}$$ $$=-1+(-1)+1+1$$ $$=0$$． 综上原式$$=0$$，故选$$\\text{A}$$． " ]

[ "1999年第10届希望杯初二竞赛第2试第3题" ]

[ [ { "aoVal": "A", "content": "$$2x-1$$ " } ], [ { "aoVal": "B", "content": "$$-2x+1$$ " } ], [ { "aoVal": "C", "content": "$$-5$$ " } ], [ { "aoVal": "D", "content": "$$1$$ " } ] ]

[ "课内体系->知识点->式->二次根式->二次根式的运算->二次根式的加减以及混合运算" ]

[ "由$$x\\textgreater\\sqrt{2}x+1$$所以$$(\\sqrt{2}-1)x\\textless{}-1$$， 所以$$x\\textless{}-\\frac{1}{\\sqrt{2}-1}=-(\\sqrt{2}+1)\\textless{}-2$$， 所以$$x\\textless{}-2$$，$$x+2\\textless{}0$$， 所以原式$$=-(x+2)+(x-3)=-5$$． 故选$$\\text{C}$$． " ]

[ "初二下学期单元测试《二次根式》二次根式的运算第43题", "2002年第19届全国初中数学联赛竞赛第1题7分" ]

[ [ { "aoVal": "A", "content": "$$a\\textless{}b\\textless{}c$$ " } ], [ { "aoVal": "B", "content": "$$a\\textless{}c\\textless{}b$$ " } ], [ { "aoVal": "C", "content": "$$b\\textless{}a\\textless{}c$$ " } ], [ { "aoVal": "D", "content": "$$b\\textless{}c\\textless{}a$$ " } ] ]

[ "课内体系->知识点->式->二次根式->二次根式的运算->二次根式比较大小" ]

[ "∵$$a-b$$ $$=\\sqrt{2}-1-\\left( 2\\sqrt{2}-\\sqrt{6} \\right)$$ $$=\\sqrt{6}-\\left( 1+\\sqrt{2} \\right)$$ $$\\approx 2.449-2.414\\textgreater0$$， ∴$$a\\textgreater b$$； ∵$$a-c$$ $$=\\sqrt{2}-1-\\left( \\sqrt{6}-2 \\right)$$ $$=\\sqrt{2}+1-\\sqrt{6}$$ $$\\approx 2.414-2.449\\textless{}0$$， ∴$$a\\textless{}c$$； 于是$$b\\textless{}a\\textless{}c$$． " ]

[ "2000年第11届希望杯初二竞赛第2试第4题" ]

[ [ { "aoVal": "A", "content": "$$-3$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$3$$或$$-1$$ " } ], [ { "aoVal": "D", "content": "$$-3$$或$$1$$ " } ] ]

[ "竞赛->知识点->数与式->数的运算->有理数运算问题" ]

[ "因为 $$a+b+c=0$$，$$abc\\textgreater0$$， 所以$$a$$，$$b$$，$$c$$中有两个负数． 由$$a+b+c=0$$，可得$$b+c=-a$$，$$c+a=-b$$，$$a+b=-c$$， 所以原式变形为$$\\frac{-a}{\\left\\textbar{} a \\right\\textbar}+\\frac{-b}{\\left\\textbar{} b \\right\\textbar}+\\frac{-c}{\\left\\textbar{} c \\right\\textbar}$$ ． 而$$\\frac{-a}{\\left\\textbar{} a \\right\\textbar}$$ ， $$\\frac{-b}{\\left\\textbar{} b \\right\\textbar}$$ ，$$\\frac{-c}{\\left\\textbar{} c \\right\\textbar}$$ ， 必有两个式子的值为$$1$$，另一个式子的值为$$-1$$， 所以原式的值为$$1$$． ", "\n<p>特殊值法．</p>\n<p>由$$a+b+c=0$$，$$abc&gt;0$$，</p>\n<p>可设$$a=2$$，$$b=c=-1$$，</p>\n<p>将其代入原式，得原式$$=\\frac{-2}{2}+\\frac{1}{1}+\\frac{1}{1}=1$$ ．</p>\n<p>故选$$\\text{B}$$．</p>" ]

[ "1998年竞赛（全国初中数学竞赛）第1题6分", "2015~2016学年浙江温州乐清市乐清市育英寄宿学校初三上学期期中实验A班第2题5分" ]

[ [ { "aoVal": "A", "content": "$$ab\\textgreater bc$$ " } ], [ { "aoVal": "B", "content": "$$a+b\\textgreater b+c$$ " } ], [ { "aoVal": "C", "content": "$$a-b\\textgreater b-c$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{a}{c}\\textgreater\\frac{b}{c}$$ " } ] ]

[ "课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质" ]

[ "根据不等式性质： ∵$$a\\textgreater b\\textgreater c$$， ∴$$a+b\\textgreater b+c$$，故$$\\text{B}$$正确， ∵$$a$$、$$b$$、$$c$$的符号不确定，∴$$\\text{A}$$、$$\\text{C}$$、$$\\text{D}$$不成立， 故选$$\\text{B}$$． " ]

[ "2005年第16届希望杯初二竞赛复赛第10题4分", "初一下学期其它第17题" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$2$$或$$14$$ " } ], [ { "aoVal": "D", "content": "$$14$$或$$17$$ " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->二元一次方程（组）->解二元一次方程组->解含绝对值的方程组" ]

[ "由②可知$$\\left\\textbar{} x-y \\right\\textbar$$、$$\\left\\textbar{} y-z \\right\\textbar$$、$$\\left\\textbar{} z-x \\right\\textbar$$中必须有一个为$$0$$，只有$$x-y=0$$，$$x=y$$，代入②中得$$\\left\\textbar{} z-x \\right\\textbar=1$$ $$z=x+1$$将$$x=y$$，$$z=x+1$$代入①中，得$$\\left\\textbar{} x \\right\\textbar+\\left\\textbar{} 2x+1 \\right\\textbar=2$$，$$\\left\\textbar{} x \\right\\textbar=0$$，$$1$$，$$2$$．经检验$$\\left\\textbar{} x \\right\\textbar=1$$，依据题意得． " ]

[ "2005年第16届希望杯初二竞赛初赛第2题4分" ]

[ [ { "aoVal": "A", "content": "$$-1$$ " } ], [ { "aoVal": "B", "content": "$$0$$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "$$-2$$ " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式->一元一次不等式的解集", "课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式->解一元一次不等式", "课内体系->知识点->方程与不等式->不等式（组）->含参不等式->由不等式（组）的解集求参数的范围" ]

[ "因为$$x=3$$满足不等式$$mx+2\\textless{}1-4m$$，所以$$3m+2\\textless{}1-4m$$， 解得$$m\\textless{}-\\frac{1}{7}$$， 又$$m$$为整数，所以$$m$$的最大值是$$-1$$． 故选$$\\text{A}$$． " ]

[ "1997年第14届全国初中数学联赛竞赛第1题" ]

[ [ { "aoVal": "A", "content": "$$1$$个 " } ], [ { "aoVal": "B", "content": "$$2$$个 " } ], [ { "aoVal": "C", "content": "$$3$$个 " } ], [ { "aoVal": "D", "content": "$$4$$个 " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->四边形->特殊平行四边形->正方形->正方形的判定" ]

[ "($$1$$)$$-1$$的倒数也是他本身，所以这个命题不对． ($$2$$)对角线互要垂直且相等的平行四边形才是正方形，所以这个命题也不对． ($$3$$)$${{a}^{2}}$$的平方根是$$\\pm \\left\\textbar{} a \\right\\textbar$$，这是对的． ($$4$$)大于直角的角一定是钝角，钝角是$$90-180$$度之间的角． 故选$$\\text{C}$$． " ]

[ "2011年第22届全国希望杯初二竞赛复赛第3题4分" ]

[ [ { "aoVal": "A", "content": "都是锐角 " } ], [ { "aoVal": "B", "content": "都是直角 " } ], [ { "aoVal": "C", "content": "都是钝角 " } ], [ { "aoVal": "D", "content": "有三个是直角，另一个是锐角或钝角 " } ] ]

[ "课内体系->知识点->三角形->三角形及多边形->多边形->多边形的内角和定理", "课内体系->知识点->三角形->三角形及多边形->多边形->正多边形->求正多边形的内角" ]

[ "凸四边形的四个内角的和是$$360{}^{}\\circ $$． 若四个内角都是锐角，则内角和小于$$360{}^{}\\circ $$； 若四个内角都是钝角，则内角和大于$$360{}^{}\\circ $$； 若有三个是直角，另一个是锐角或钝角，则内角和不为$$360{}^{}\\circ $$； 若四个内角都是直角，则内角和等于$$360{}^{}\\circ $$，这时可以构成凸四边形，符合题意． " ]

[ "2020~2021学年四川德阳初一上学期期末第11题7分", "2000年第11届希望杯初一竞赛第8题" ]

[ [ { "aoVal": "A", "content": "$$a\\textgreater c\\textgreater b\\textgreater d$$ " } ], [ { "aoVal": "B", "content": "$$b\\textgreater d\\textgreater a\\textgreater c$$ " } ], [ { "aoVal": "C", "content": "$$c\\textgreater a\\textgreater b\\textgreater d$$ " } ], [ { "aoVal": "D", "content": "$$d\\textgreater b\\textgreater a\\textgreater c$$ " } ] ]

[ "课内体系->知识点->数->有理数->有理数比较大小->有理数比较大小-利用有理数正负性" ]

[ "由$$\\frac{1}{a-1997}=\\frac{1}{b+1998}=\\frac{1}{c-1999}=\\frac{1}{d+2000}$$， 可知$$a-1997=b+1998=c-1999=d+2000$$， 由这个连等式可得：$$a\\textgreater b$$，$$a\\textless{}c$$，$$a\\textgreater d$$；$$b\\textless{}c$$，$$b\\textgreater d$$，$$c\\textless{}d$$， 由此可得$$c\\textgreater a\\textgreater b\\textgreater d$$， 故选$$\\text{C}$$． " ]

[ "2018年第29届希望杯初一竞赛初赛第16题4分" ]

[ [ { "aoVal": "A", "content": "$$117$$ " } ], [ { "aoVal": "B", "content": "$$99$$ " } ], [ { "aoVal": "C", "content": "$$96$$ " } ], [ { "aoVal": "D", "content": "$$84$$ " } ], [ { "aoVal": "E", "content": "$$48$$ " } ] ]

[ "课内体系->能力->运算能力", "课内体系->能力->分析和解决问题能力", "课内体系->知识点->几何图形初步->几何图形->立体图形与平面图形->正方体堆积图形的问题" ]

[ "设该长方体的棱长分别为$$a$$厘米，$$b$$厘米，$$c$$厘米，$$(a\\geqslant b\\geqslant c)$$， 则由题意得$$(a-2)(b-2)(c-2)=11$$， ∵$$11$$为质数， ∴$$a-2=11$$，$$b-2=1$$，$$c-2=1$$， ∴$$a=13$$，$$b=3$$，$$c=3$$， ∴长方体的体积$$V=abc=117$$立方厘米． " ]

[ "1994年第5届全国希望杯初一竞赛初赛第3题" ]

[ [ { "aoVal": "A", "content": "$$4\\frac{1}{3}$$ " } ], [ { "aoVal": "B", "content": "$$-\\frac{3}{13}$$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "$$-1$$ " } ] ]

[ "课内体系->知识点->数->有理数->倒数与负倒数->负倒数的定义" ]

[ "$$\\frac{\\left\\textbar{} 1-9 \\right\\textbar-\\left\\textbar{} 9-4 \\right\\textbar}{1-9-9+4}=\\frac{8-5}{-13}=-\\frac{3}{13}$$，其负倒数为$$\\frac{13}{3}=4\\frac{1}{3}$$． " ]

[ "1991年第2届全国希望杯初一竞赛复赛第1题" ]

[ [ { "aoVal": "A", "content": "$$p\\geqslant q\\geqslant a\\textgreater b$$ " } ], [ { "aoVal": "B", "content": "$$q\\geqslant a\\textgreater b\\geqslant p$$ " } ], [ { "aoVal": "C", "content": "$$q\\geqslant p\\geqslant a\\textgreater b$$ " } ], [ { "aoVal": "D", "content": "$$p\\geqslant a\\textgreater b\\geqslant q$$ " } ] ]

[ "课内体系->知识点->数->有理数->有理数比较大小", "竞赛->知识点->数论->整除->因数与倍数" ]

[ "两个自然数的最小公倍数一定不小于两数中较大者． 两个自然数的最大公约数一定不大于两数中较小者． 所以$$q\\geqslant a\\textgreater b\\geqslant p$$． 选$$\\text{B}$$． " ]

[ "2013年第24届全国希望杯初二竞赛初赛第7题4分" ]

[ [ { "aoVal": "A", "content": "$$100$$个 " } ], [ { "aoVal": "B", "content": "$$101$$个 " } ], [ { "aoVal": "C", "content": "$$201$$个 " } ], [ { "aoVal": "D", "content": "$$203$$个 " } ] ]

[ "课内体系->知识点->数->有理数->绝对值->已知范围化简绝对值", "课内体系->能力->运算能力" ]

[ "此题暂无解析 " ]

[ "2007年全美数学竞赛（AMC）竞赛" ]

[ [ { "aoVal": "A", "content": "$$\\frac 27$$ " } ], [ { "aoVal": "B", "content": "$$\\frac 38$$ " } ], [ { "aoVal": "C", "content": "$$\\frac 12$$ " } ], [ { "aoVal": "D", "content": "$$\\frac 47$$ " } ], [ { "aoVal": "E", "content": "$$\\frac 58$$ " } ] ]

[ "美国AMC8->Knowledge Point->Counting, Probability and Statistics->Classical Probability", "课内体系->知识点->统计与概率" ]

[ "从写有$$A$$、$$B$$、$$C$$、$$D$$的四张红牌和$$A$$、$$B$$、$$C$$、$$D$$的四张绿牌中抽取两张．获胜对子为两张相同花色或相同数字的．则抽到获胜对子的概率是？ 有四种方法可以选择同一个字母的一对，并且$$2\\left( \\left( \\begin{matrix}4 2 \\end{matrix} \\right) \\right)=12$$方法去选择一对相同颜色．一共有$$\\left( \\begin{matrix}8 2 \\end{matrix} \\right)=28$$方法去选择一对，所以概率是$$\\frac{4+12}{28}=\\frac{4}{7}$$． 故选$$\\text{D}$$． There are $$4$$ ways of choosing a winning pair of the same letter, and $$2\\left( \\left( \\begin{matrix}4 2 \\end{matrix} \\right) \\right)=12$$ ways to choose a pair of the same color. There\\textquotesingle s a total of $$\\left( \\begin{matrix}8 2 \\end{matrix} \\right)=28$$ ways to choose a pair, so the probability is $$\\frac{4+12}{28}=\\frac{4}{7}$$. " ]

[ "2019年第1届广东深圳罗湖区深圳中学初中部初一竞赛（凤凰木杯）第6题3分" ]

[ [ { "aoVal": "A", "content": "$$44$$ " } ], [ { "aoVal": "B", "content": "$$45$$ " } ], [ { "aoVal": "C", "content": "$$46$$ " } ], [ { "aoVal": "D", "content": "$$49$$ " } ] ]

[ "课内体系->知识点->式->整式的乘除->乘法公式->平方差公式的计算", "课内体系->能力->运算能力" ]

[ "∵$$a_{{}}^{2}-b_{{}}^{2}+9-2018$$， ∴$$a_{{}}^{2}-b_{{}}^{2}=2009$$， ∴$$(a+b)(a-b)=2009$$， ∵$$2009=7\\times 7\\times 41=41\\times 49$$， ∵$$a\\cdot b$$都是正整数， ∴$$a+b$$为正整数，且$$a+b\\textgreater a-b$$， ∴$$a+b=49$$， 故选$$\\text{D}$$． " ]

[ "2011年第22届全国希望杯初一竞赛初赛第3题4分" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "课内体系->知识点->数->有理数->绝对值->绝对值综合", "课内体系->知识点->数->有理数->绝对值->绝对值的几何意义", "课内体系->知识点->数->有理数->绝对值->绝对值的非负性" ]

[ "①令$$a=-2$$，$$b=2$$，则①错误； ②令$$a=2$$，$$b=-3$$，则②错误； ③若$$a=-b$$，则$${{\\left( -a \\right)}^{2}}={{\\left[ -(-b) \\right]}^{2}}={{b}^{2}}$$，则③正确； ④令$$a=-1$$，$$b=-2$$，则④错误． 故正确的判断的个数是$$1$$． " ]

[ "2020~2021学年四川成都金牛区成都外国语学校初二上学期期中模拟第8题3分", "2018~2019学年广东深圳罗湖区深圳中学初中部初一下学期期中（竞赛班）第5题2分" ]

[ [ { "aoVal": "A", "content": "$$x\\textgreater y$$ " } ], [ { "aoVal": "B", "content": "$$x\\geqslant y$$ " } ], [ { "aoVal": "C", "content": "$$x\\textless{}y$$ " } ], [ { "aoVal": "D", "content": "$$x=y$$ " } ] ]

[ "课内体系->知识点->数->实数->无理数有关的计算->无理数大小的比较" ]

[ "∵$$x=\\frac{4}{\\sqrt{5}+3}=\\frac{4(\\sqrt{5}-3)}{(\\sqrt{5}+3)(\\sqrt{5}-3)}=-(\\sqrt{5}-3)=3-\\sqrt{5}\\textgreater0$$， $$y=\\sqrt{5}-3\\textless{}0$$， ∴$$x\\textgreater y$$， 故选$$\\text{A}$$． " ]

[ "2007年竞赛第2题6分" ]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "竞赛->知识点->组合->计数问题-枚举法" ]

[ "用枚举法： 红球个数 白球个数 黑球个数 种数 $$5$$ $$2$$，$$3$$，$$4$$，$$5$$ $$3$$，$$2$$，$$1$$，$$0$$ $$4$$ $$4$$ $$3$$，$$4$$，$$5$$，$$6$$ $$3$$，$$2$$，$$1$$，$$0$$ $$4$$ $$3$$ $$4$$，$$5$$，$$6$$，$$7$$ $$3$$，$$2$$，$$1$$，$$0$$ $$4$$ $$2$$ $$5$$，$$6$$，$$7$$，$$8$$ $$3$$，$$2$$，$$1$$，$$0$$ $$4$$ 所以共$$16$$种．故选$$B$$． " ]

[ "1995年第6届全国希望杯初一竞赛初赛第1题" ]

[ [ { "aoVal": "A", "content": "$$19$$ " } ], [ { "aoVal": "B", "content": "$$-19$$ " } ], [ { "aoVal": "C", "content": "$$0$$ " } ], [ { "aoVal": "D", "content": "$$1$$ " } ] ]

[ "课内体系->知识点->式->整式的加减->整式有关的概念->代数式->代数式的定义", "课内体系->知识点->式->整式的加减->整式有关的概念->代数式->列代数式" ]

[ "当$$a=14$$时，$$-\\frac{95}{a-19}=19$$，排除$$\\text{A}$$； 当$$a=24$$时，$$-\\frac{95}{a-19}=-19$$，排除$$\\text{B}$$； 当$$a=-76$$时，$$-\\frac{95}{a-19}=1$$，排除$$\\text{D}$$； 因此，选$$\\text{C}$$．事实上，对任意$$a\\ne 19$$，$$-\\frac{95}{a-19}$$一定不是$$0$$． " ]

[ "1993年第4届希望杯初二竞赛第14题" ]

[ [ { "aoVal": "A", "content": "$$\\pm 2\\sqrt{3}$$ " } ], [ { "aoVal": "B", "content": "$$2\\sqrt{3}$$ " } ], [ { "aoVal": "C", "content": "$$\\pm 2\\sqrt{2}$$ " } ], [ { "aoVal": "D", "content": "$$2\\sqrt{2}$$ " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->一元二次方程->根与系数的关系->一元二次方程根与系数的关系", "课内体系->知识点->方程与不等式->一元二次方程->根的判别式的应用->一元二次方程根的判别式" ]

[ "因为$$\\Delta \\geqslant 0$$，所以$$36{{(a+1)}^{2}}-36({{a}^{2}}-3)\\geqslant 0$$，所以$$a\\geqslant -2$$，又$${{x}_{1}}\\cdot {{x}_{2}}=1$$，所以$$\\frac{{{a}^{2}}-3}{9}=1$$，解得$$a=\\pm 2\\sqrt{3}$$．又由$$a\\geqslant -2$$，所以$$a=2\\sqrt{3}$$．故选$$\\text{B}$$． " ]

[ "2019~2020学年12月四川资阳雁江区资阳市雁江区第二中学初三上学期周测D卷第10题3分", "2016~2017学年3月湖北武汉武昌区武汉初级中学初二下学期月考第10题3分", "2016~2017学年9月湖北武汉武昌区武汉初级中学初二上学期月考第10题3分", "2001年第18届全国初中数学联赛竞赛第1题" ]

[ [ { "aoVal": "A", "content": "$$1999$$ " } ], [ { "aoVal": "B", "content": "$$2000$$ " } ], [ { "aoVal": "C", "content": "$$2001$$ " } ], [ { "aoVal": "D", "content": "不能确定 " } ] ]

[ "课内体系->知识点->式->二次根式->二次根式的运算->二次根式的四则混合运算", "课内体系->知识点->方程与不等式->等式与方程->等式->等式的性质->等式性质1", "课内体系->能力->运算能力" ]

[ "本题需要比较等式两边的各项，利用有理数部分等于理数部分，无理数部分等于无理数部分来求$$a$$、$$b$$、$$c$$的值，由于： $$5+2\\sqrt{6}={{\\left( \\sqrt{3}+\\sqrt{2} \\right)}^{2}}$$． 所以：$$a+b\\sqrt{2}+c\\sqrt{3}=\\sqrt{2}+\\sqrt{3}$$． 则$$a=0$$，$$b=1$$，$$c=1$$，∴$$2a+999b+1001c=2000$$． 所以选$$\\text{B}$$． " ]

[ "1992年第9届全国初中数学联赛竞赛第3题", "初一上学期其它" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ] ]

[ "课内体系->知识点->式->整式的乘除->乘法公式->完全平方公式的计算-不含分式", "课内体系->能力->运算能力" ]

[ "由$${{x}^{2}}-13x+1$$可得到$$x+\\dfrac{1}{x}=13$$，所以$${{x}^{2}}+\\dfrac{1}{{{x}^{2}}}={{(x+\\dfrac{1}{x})}^{2}}-2={{13}^{2}}-2=167$$． 同样的$${{x}^{4}}+\\dfrac{1}{{{x}^{4}}}={{({{x}^{2}}+\\dfrac{1}{{{x}^{2}}})}^{2}}-2$$，可以根据个位数字可以直接判断结果的个位数字为$$7$$． " ]

[ "1991年第2届希望杯初二竞赛第10题" ]

[ [ { "aoVal": "A", "content": "$$26$$ " } ], [ { "aoVal": "B", "content": "$$28$$ " } ], [ { "aoVal": "C", "content": "$$36$$ " } ], [ { "aoVal": "D", "content": "$$38$$ " } ] ]

[ "竞赛->知识点->数与式->数的运算->有理数运算问题" ]

[ "设这个两位数为$$\\overline{ab}$$，则$$3(a+b)+2=10a+b$$， 即$$7a=2b+2$$， 可见$$a$$只能为偶数，$$b+1$$是$$7$$的倍数． ∴这个两位数是$$28$$， 故选$$\\text{A}$$． " ]

[ "2010年第21届希望杯初二竞赛第1试第6题", "北京初二上学期单元测试《二次根式的运算》第22题" ]

[ [ { "aoVal": "A", "content": "$$p\\textgreater5$$ " } ], [ { "aoVal": "B", "content": "$$p\\textless{}5$$ " } ], [ { "aoVal": "C", "content": "$$p\\textless{}4$$ " } ], [ { "aoVal": "D", "content": "$$p=5$$ " } ] ]

[ "课内体系->能力->推理论证能力", "课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质", "课内体系->知识点->式->因式分解->公式法->利用立方和与立方差公式因式分解", "课内体系->知识点->式->整式的乘除->乘法公式->和与差的立方公式" ]

[ "∵$$a$$、$$b$$、$$c$$、$$d$$是正实数，且$$a+b+c+d=1$$， ∴$$0\\textless{}a\\textless{}1$$， ∴$$a\\textgreater{{a}^{2}}\\textgreater{{a}^{3}}$$， ∴$$7a+1=a+3a+3a+1\\textgreater{{a}^{3}}+3{{a}^{2}}+3a+1={{(a+1)}^{3}}$$， ∴$$\\sqrt[3]{7a+1}\\textgreater\\sqrt[3]{{{(a+1)}^{3}}}=a+1$$， 同理$$\\sqrt[3]{7b+1}\\textgreater b+1\\sqrt[3]{7c+1}\\textgreater c+1\\sqrt[3]{7d+1}\\textgreater d+1$$， ∴$$p=\\sqrt[3]{7a+1}+\\sqrt[3]{7b+1}+\\sqrt[3]{7c+1}+\\sqrt[3]{7d+1}$$， $$\\textgreater a+1+b+1+c+1+d+1$$， $$=a+b+c+d+4=5$$， ∴$$p\\textgreater5$$． " ]

[ "1994年第11届全国初中数学联赛竞赛第4题6分", "初一下学期其它" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$-1$$ " } ], [ { "aoVal": "C", "content": "$${{2}^{2001}}$$ " } ], [ { "aoVal": "D", "content": "$$-{{2}^{2001}}$$ " } ] ]

[ "课内体系->知识点->式->二次根式->二次根式化简求值->二次根式的化简求值——利用完全平方" ]

[ "∵$$x=\\frac{1+\\sqrt{1994}}{2}$$， ∴$${{(2x-1)}^{2}}=1994$$， 即$$4{{x}^{2}}-4x-1993=0$$， ∴$${{(4{{x}^{3}}-1997x-1994)}^{2001}}$$ $$={{[(4{{x}^{2}}-4x-1993)x+(4{{x}^{2}}-4x-1993)-1]}^{2001}}$$ $$={{(-1)}^{2001}}$$ $$=-1$$． " ]

[ "2003年第14届希望杯初二竞赛第2试第7题" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$10$$或$$9$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ] ]

[ "课内体系->能力->推理论证能力", "课内体系->知识点->三角形->三角形及多边形->多边形->多边形的内角和定理", "课内体系->知识点->三角形->三角形及多边形->多边形->求多边形的内角和" ]

[ "因为，在四边形中，两个最大的内角的和大于等于$$180{}^{}\\circ $$， 所以，凸$$n$$边形（$$n\\geqslant 4$$）中，两个最大的内角和也大于等于$$180{}^{}\\circ $$，同时小于$$360{}^{}\\circ $$， 设两个最大的内角的和为$$k{}^{}\\circ $$，则$$1100+180\\leqslant 1100+k=(n-2)\\times 180 ~\\textless{} ~1100+360$$， 所以$$7\\frac{1}{9}\\leqslant n-2 ~\\textless{} ~8\\frac{1}{9}$$，$$9\\frac{1}{9}\\leqslant n ~\\textless{} ~10\\frac{1}{9}$$，$$n=10$$． 故选$$\\text{D}$$． " ]

[ "2006年第17届希望杯初二竞赛第2试第5题", "初二其它" ]

[ [ { "aoVal": "A", "content": "是完全平方数，还是奇数 " } ], [ { "aoVal": "B", "content": "是完全平方数，还是偶数 " } ], [ { "aoVal": "C", "content": "不是完全平方数，但是奇数 " } ], [ { "aoVal": "D", "content": "不是完全平方数，但是偶数 " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->式->因式分解->十字相乘法->二次项系数为±1的十字相乘" ]

[ "$$m={{2006}^{2}}+{{2006}^{2}}\\times {{2007}^{2}}+{{2007}^{2}}$$ $$={{2006}^{2}}+{{2006}^{2}}{{\\left( 2006+1 \\right)}^{2}}+{{\\left( 2006+1 \\right)}^{2}}$$ $$={{2006}^{4}}+2\\times {{2006}^{3}}+3\\times {{2006}^{2}}+2\\times 2006+1$$ $$={{\\left( {{2006}^{2}}+2006+1 \\right)}^{2}}$$ ∴$$m$$是奇数． 故选$$\\text{A}$$． " ]

[ "1990年第1届希望杯初二竞赛第7题" ]

[ [ { "aoVal": "A", "content": "$$\\sqrt{-a}$$ " } ], [ { "aoVal": "B", "content": "$$-\\sqrt{a}$$ " } ], [ { "aoVal": "C", "content": "$$-\\sqrt{-a}$$ " } ], [ { "aoVal": "D", "content": "$$\\sqrt{a}$$ " } ] ]

[ "课内体系->知识点->式->二次根式->二次根式的基础->最简二次根式" ]

[ "因为$$a\\textless{}0$$，所以$$a\\sqrt{-\\frac{1}{a}}=-\\sqrt{{{(-a)}^{2}}\\frac{1}{-a}}=-\\sqrt{-a}$$． 故选$$\\text{C}$$． " ]

[ "2016年第27届全国希望杯初一竞赛复赛第3题4分" ]

[ [ { "aoVal": "A", "content": "$$16$$元 " } ], [ { "aoVal": "B", "content": "$$14.8$$元 " } ], [ { "aoVal": "C", "content": "$$11.5$$元 " } ], [ { "aoVal": "D", "content": "$$10.7$$元 " } ] ]

[ "课内体系->知识点->方程与不等式->二元一次方程（组）->二元一次方程组应用题->二元一次方程组的经济问题", "课内体系->知识点->方程与不等式->二元一次方程（组）->解二元一次方程组->加减消元法解二元一次方程组" ]

[ "设$$1$$斤苹果$$x$$元，$$1$$斤西瓜$$y$$元，$$1$$斤橙子$$z$$元，$$1$$斤葡萄$$w$$元． 由题意，得$$\\left { \\begin{matrix}x+4y+2z+w=27.6\\&① 2x+6y+2z+2w=32.2 \\&② \\end{matrix} \\right.\\begin{matrix} \\end{matrix}$$， $$2\\times $$①$$-$$②，得$$2y+2z=23$$， ∴$$y+z=11.5$$． 答：买$$1$$斤西瓜和$$1$$斤橙子需付$$11.5$$元． " ]

[ "1993年第4届希望杯初二竞赛第2题" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "课内体系->知识点->数->实数->无理数有关的计算->无理数的定义", "课内体系->知识点->数->实数->立方根->平方根和立方根综合", "课内体系->知识点->数->实数->平方根", "课内体系->知识点->几何图形初步->命题与证明" ]

[ "命题①，③是正确的，②，④不正确． 故选$$\\text{B}$$． " ]

[ "2001年竞赛（全国初中数学竞赛）第3题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{123}{22}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{125}{22}$$或$$2$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{125}{22}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{123}{22}$$或$$2$$ " } ] ]

[ "课内体系->知识点->式->分式->分式化简求值->分式化简求值-条件化简求值", "课内体系->思想->分类讨论思想" ]

[ "有两种情况： ①若$$a=b$$，则$$\\frac{b}{a}+\\frac{a}{b}=2$$； ②若$$a\\ne b$$，根据题意，$$a$$、$$b$$是方程$${{x}^{2}}-13x+m=0$$的根， 则$$a+b=13$$，因为$$a$$，$$b$$是质数且和为奇数，所以两数分别为$$2$$和$$11$$．此时$$\\frac{b}{a}+\\frac{a}{b}=\\frac{2}{11}+\\frac{11}{2}=\\frac{125}{22}$$． 方法二：两式相减，消$$m$$，$${{a}^{2}}-{{b}^{2}}-13a+13b=0$$，$$\\left( a-b \\right)\\left( a+b-13 \\right)=0$$，所以有$$a=b$$或$$a+b=13$$. " ]

[ "2020~2021学年10月湖南长沙天心区长郡外国语实验中学初一上学期周测A卷", "1996年第7届希望杯初二竞赛第2题4分" ]

[ [ { "aoVal": "A", "content": "$$a+b ~\\textless{} ~a-b ~\\textless{} ~a-1 ~\\textless{} ~a+1$$ " } ], [ { "aoVal": "B", "content": "$$a+1\\textgreater a+b\\textgreater a-b\\textgreater a-1$$ " } ], [ { "aoVal": "C", "content": "$$a-1 ~\\textless{} ~a+b ~\\textless{} ~a-b ~\\textless{} ~a+1$$ " } ], [ { "aoVal": "D", "content": "$$a+b\\textgreater a-b\\textgreater a+1\\textgreater a-1$$ " } ] ]

[ "课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质" ]

[ "因为$$-1 ~\\textless{} ~b ~\\textless{} ~a ~\\textless{} ~0$$，所以$$a+b ~\\textless{} ~a-b$$． 因为$$b\\textgreater-1$$，所以$$a-1 ~\\textless{} ~a+b$$． 又因为$$-b ~\\textless{} ~1$$，所以$$a-b ~\\textless{} ~a+1$$． 综上得$$a-1 ~\\textless{} ~a+b ~\\textless{} ~a-b ~\\textless{} ~a+1$$． 故选$$\\text{C}$$． " ]

[ "1994年第5届全国希望杯初一竞赛初赛第8题" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ] ]

[ "课内体系->知识点->方程与不等式->等式与方程->等式->等式的定义", "课内体系->知识点->方程与不等式->二元一次方程（组）->二元一次方程->二元一次方程的解" ]

[ "由$$x:y=3:2$$得$$x=1.5y$$， 代入$$x+3y=27$$得$$4.5y=27$$， 于是$$y=6$$，$$x=9$$， 所以$$x$$，$$y$$中较小的那个数是$$6$$． " ]

[ "2007年第12届华杯赛初一竞赛初赛第6题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{3a+2}{5}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2-a}{3}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{3a+1}{6}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{5a-2}{7}$$ " } ] ]

[ "课内体系->知识点->数->有理数->有理数基础运算->有理数减法->有理数加减混合运算", "课内体系->能力->运算能力" ]

[ "取$$a=1$$，$$\\frac{3a+2}{5}$$，否定$$\\text{A}$$． 取$$a=5$$，$$\\frac{2-a}{3}=-1$$，否定$$\\text{B}$$． 取$$a=-1$$，$$\\frac{5a-2}{7}=-1$$，否定$$\\text{D}$$． 又当$$a$$是整数时，$$3a+1$$表示被$$3$$除余$$1$$的整数，故不可能是$$6$$的倍数． 所以$$\\frac{3a+1}{6}$$不可能得整数值． 故选$$\\text{C}$$． " ]

[ "2022~2023学年浙江宁波初三月考（六校强基竞赛）第6题" ]

[ [ { "aoVal": "A", "content": "当$n\\textgreater{} 0$时，$m\\textless{} x_{1}$ " } ], [ { "aoVal": "B", "content": "当$n\\textgreater{} 0$时，$m\\textgreater{} x_{2}$ " } ], [ { "aoVal": "C", "content": "当$n\\textless{} 0$时，$m\\textless{} 0$ " } ], [ { "aoVal": "D", "content": "当$n\\textless{} 0$时，$x_{1}\\textless{} m\\textless{} x_{2}$ " } ] ]

[ "课内体系->知识点->函数->二次函数" ]

[ "\\hfill\\break 【分析】\\\\ 根据二次函数的图象与性质可进行排除选项．\\\\ 【详解】\\\\ 解：由二次函数$y=x^{2}+2x+c$可知开口向上，对称轴为直线$x=-1$，\\\\ 当$x\\textless{} -1$时，$y$随$x$的增大而减小，当$x\\textgreater{} -1$时，\\emph{y}随\\emph{x}的增大而增大；\\\\ ∵$A\\left(x_{1},0\\right)$，$B\\left(x_{2},0\\right)$是二次函数与\\emph{x}轴的交点，点$P\\left(m,n\\right)$是图象上的一点，\\\\ ∴当$n\\textgreater{} 0$时，则$m\\textless{} x_{1}$或$m\\textgreater{} x_{2}$；故$\\mathrm{A}$、$\\mathrm{B}$选项错误；\\\\ 当$n\\textless{} 0$时，则$x_{1}\\textless{} m\\textless{} x_{2}$，故$\\mathrm{D}$正确；当$x_{2}\\textgreater{} 0$且$n\\textless{} 0$时，此时有可能$m\\textgreater{} 0$，故$\\mathrm{C}$错误；\\\\ 故选$\\mathrm{D}$．\\\\ 【点睛】\\\\ 本题主要考查二次函数的图象与性质，熟练掌握二次函数的图象与性质是解题的关键． " ]

[ "2016年全美数学竞赛（AMC）竞赛" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$7$$ " } ] ]

[ "课内体系->知识点->式->因式分解->十字相乘法->二次项系数为±1的十字相乘", "美国AMC8->Knowledge Point->Number Theory->Remainder Problems->Remainder Characteristics" ]

[ "数字$$N$$是个两位数． $$N$$除以$$9$$，余数为$$1$$． $$N$$除以$$10$$，余数为$$3$$． 则$$N$$除以$$11$$时，余数为几？ 从第二个子弹开始，我们知道第二个数字必须是$$3$$，因为除以$$9$$还有一个余数$$1$$，$$9$$的倍数必须以$$2$$结束，我们找到： $$9\\left( 1 \\right)=9$$ $$9\\left( 2 \\right)=18$$ $$9\\left( 3 \\right)=27$$ $$9\\left( 4 \\right)=36$$ $$9\\left( 5 \\right)=45$$ $$9\\left( 6 \\right)=54$$ $$9\\left( 7 \\right)=63$$ $$9\\left( 8 \\right)=72$$ 这个上数字$$71+1=73$$都满足条件，我们减去$$11$$的小于$$73$$的最大倍数得到剩余的，也就是$$73-11\\left( 6 \\right)=73-66=7$$． 故选$$\\text{E}$$． From the second bullet point, we know that the second digit must be $$3$$. Because there is a remainder of $$1$$ when it is divided by $$9$$ the multiple of $$9$$ must end in a $$2$$. We now look for this one: $$9(1) = 9$$ $$9(2) =18$$ $$9(3) = 27$$ $$9(4) = 36$$ $$9(5) = 45$$ $$9(6) =54$$ $$9(8) = 72$$ The number $$72 +1= 73$$ satisfies both conditions. We subtract the biggest multiple of $$11$$ less than $$73$$ to get the remainder. Thus, $$73- 11(6) = 73 - 66 =\\boxed{(\\rm E) 7 }$$. " ]

[ "1992年第3届全国希望杯初一竞赛初赛第8题", "2019~2020学年辽宁沈阳皇姑区沈阳市33中学初一上学期期中第5题2分" ]

[ [ { "aoVal": "A", "content": "$$4a-b$$ " } ], [ { "aoVal": "B", "content": "$$b-a$$ " } ], [ { "aoVal": "C", "content": "$$a-9b$$ " } ], [ { "aoVal": "D", "content": "$$7b$$ " } ] ]

[ "课内体系->知识点->式->整式的加减->整式的加减运算->整式加减", "课内体系->能力->运算能力" ]

[ "$$(2a+5b)-\\frac{1}{2}(4a-4b)=2a+5b-2a+2b=7b$$，选$$\\text{D}$$． " ]

[ "2005年第16届希望杯初二竞赛复赛第1题4分", "其它" ]

[ [ { "aoVal": "A", "content": "$$m$$一定是奇数 " } ], [ { "aoVal": "B", "content": "$$m$$一定是偶数 " } ], [ { "aoVal": "C", "content": "只有$$a$$,$$b$$当均为偶数时，$$m$$是偶数 " } ], [ { "aoVal": "D", "content": "只有$$a$$,$$b$$当一个为偶数、另一个为奇数时，$$m$$是偶数 ~ " } ] ]

[ "竞赛->知识点->数论->同余->奇数与偶数", "课内体系->知识点->数->有理数->有理数基础运算->有理数除法->有理数四则混合运算" ]

[ "$$m=ab(a+b)$$ 若$$a$$，$$b$$中有一个为偶数，$$m$$必为偶数． 若$$a$$，$$b$$均为奇数，$$a+b$$为偶数，$$m$$必为偶数． " ]

[ "2016年第27届全国希望杯初一竞赛初赛第5题4分" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$1024$$ " } ], [ { "aoVal": "C", "content": "$$2104$$ " } ], [ { "aoVal": "D", "content": "$$2016$$ " } ] ]

[ "课内体系->能力->推理论证能力", "课内体系->能力->运算能力", "课内体系->知识点->数->有理数->有理数基础运算->有理数除法->有理数四则混合运算", "竞赛->知识点->数论->整除->因数与倍数" ]

[ "设$$a$$，$$b$$的最大公约数是$$c$$， 则它们的最小公倍数为$$c\\cdot \\frac{a}{c}\\cdot \\frac{b}{c}=\\frac{ab}{c}$$， 由题意，得$$c+\\frac{ab}{c}=ab$$， ∴$$ab=\\frac{{{c}^{2}}}{c-1}$$． 若$$c$$为奇数，则$${{c}^{2}}$$为奇数，$$c-1$$为偶数，$$\\frac{{{c}^{2}}}{c-1}$$不是整数，舍去． 若$$c$$为偶数，则$${{c}^{2}}$$为偶数，$$c-1$$为奇数，当且仅当$$c=2$$时，$$\\frac{{{c}^{2}}}{c-1}$$为整数． ∴$$ab=\\frac{{{2}^{2}}}{2-1}=4$$． 又∵$$a$$，$$b$$均为正整数，且它们的最大公约数为$$2$$， ∴$$a=b=2$$， ∴$${{\\left( \\frac{{{a}^{2}}{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}} \\right)}^{10}}={{\\left( \\frac{16}{8} \\right)}^{10}}={{2}^{10}}=1024$$． " ]

[ "2009年第20届希望杯初二竞赛第1试第10题" ]

[ [ { "aoVal": "A", "content": "$$0$$个 " } ], [ { "aoVal": "B", "content": "$$1$$个 " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "无穷多个 " } ] ]

[ "课内体系->知识点->式->整式的加减->整式的加减运算->整式加减" ]

[ "当$$x=3$$时，$$a+3b+9c=2008$$，① 当$$x=7$$时，$$a+7b+49c=2009$$， ② ②$$-$$①得，$$4b+40c=1$$． 当$$b$$，$$c$$都是整数时，上式左边总为偶数，不可能等于$$1$$，所以不存在这样的代数式． 故选$$\\text{A}$$． " ]

[ "2014年第25届全国希望杯初一竞赛复赛第10题4分" ]

[ [ { "aoVal": "A", "content": "$$0.9$$ " } ], [ { "aoVal": "B", "content": "$$1.8$$ " } ], [ { "aoVal": "C", "content": "$$3.6$$ " } ], [ { "aoVal": "D", "content": "$$7.2$$ " } ] ]

[ "课内体系->能力->运算能力", "课内体系->能力->分析和解决问题能力", "课内体系->知识点->数->有理数->有理数与实际问题" ]

[ "由已知条件可得，$$1$$号阀门每分钟注入$$\\frac{1}{18}$$池水，而$$2$$号阀门每分钟放出$$\\frac{1}{24}$$池水． 到$$A$$池深$$0.4$$米时，$$A$$池中正好留存了$$\\frac{1}{3}$$池水， 则注水时间为$$\\frac{1}{3}\\div \\left( \\frac{1}{18}-\\frac{1}{24} \\right)=24$$（分钟）， 故同时关闭阀门时，恰好放了$$24$$分钟水，正好把$$B$$池放满， 故$$B$$水池中有水$$3\\times 2\\times 1.2=7.2$$（立方米）． " ]

[ "1999年第10届希望杯初一竞赛第10题" ]

[ [ { "aoVal": "A", "content": "$$5 \\%$$ " } ], [ { "aoVal": "B", "content": "$$10 \\%$$ " } ], [ { "aoVal": "C", "content": "$$15 \\%$$ " } ], [ { "aoVal": "D", "content": "$$20 \\%$$ " } ] ]

[ "课内体系->知识点->方程与不等式->一元二次方程->一元二次方程与实际问题->一元二次方程的增长率问题" ]

[ "设$$1998$$年的产值为$$1$$，则$$1999$$年产值为$$1+x$$，$$2000$$年产值为$$(1+x)+(1+x)x={{(1+x)}^{2}}$$．依题意，应当$${{(1+x)}^{2}}=1.44$$． 因为$${{(\\pm 1.2)}^{2}}=1.44$$，且$$x\\textgreater0$$， 所以$$1+x=1.2$$，$$x=0.2=20 \\%$$． 故选$$\\text{D}$$． " ]

[ "2001年第6届华杯赛初一竞赛决赛第19题" ]

[ [ { "aoVal": "A", "content": "$$1173$$ " } ], [ { "aoVal": "B", "content": "$$1731$$ " } ], [ { "aoVal": "C", "content": "$$7311$$ " } ], [ { "aoVal": "D", "content": "$$3117$$ " } ], [ { "aoVal": "E", "content": "$$1713$$ " } ] ]

[ "竞赛->知识点->组合->操作与游戏", "课内体系->知识点->综合与实践->规律探究与程序框图->规律探究->数列找规律->数列找规律-具有周期规律的数列" ]

[ "第$$1$$次操作得数字串$$711131737$$；第$$2$$次操作得数字串$$11133173$$；第$$3$$次操作得数字串$$111731$$；第$$4$$次操作得数字串$$1173$$；第$$5$$次操作得数字串$$1731$$；第$$6$$次操作得数字串$$7311$$；第$$7$$次操作得数字串$$3117$$；第$$8$$次操作得数字串$$1173$$，以下以$$4$$为周期循环，即$$4k$$次操作所得数字串均为$$1173$$．$$1996=4\\times 499$$，所以第$$1996$$次操作得数字串$$1173$$，因此，第$$1997$$次操作得数字串$$1731$$． " ]

[ "1998年第9届希望杯初一竞赛第4题" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$3$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "竞赛->知识点->几何图形初步->角->角的基础问题" ]

[ "由$$3-4=-1$$，知命题①不真； $$3a{{b}^{2}}$$与$$5a{{b}^{2}}$$是同类项，但数字系数不同，③不真； 由于两条平行线被第三条直线所截，同旁内角之和为$$180{}^{}\\circ $$，但它们并不互为邻补角．命题④不真． 易知，两个整式的和仍是整式是真命题． 所以只有$$1$$个真命题，选$$\\text{A}$$． " ]

[ "2006年第17届希望杯初二竞赛第2试第1题" ]

[ [ { "aoVal": "A", "content": "$$\\sqrt{2.5}$$和$$2\\sqrt{0.5}$$ " } ], [ { "aoVal": "B", "content": "$$3a\\sqrt{a}$$和$$3b\\sqrt{b}$$ " } ], [ { "aoVal": "C", "content": "$$\\sqrt{{{a}^{2}}b}$$和$$\\sqrt{a{{b}^{2}}}$$ " } ], [ { "aoVal": "D", "content": "$$\\sqrt{a{{b}^{7}}{{c}^{3}}}$$和$$\\sqrt{\\frac{{{c}^{3}}}{ab}}$$ " } ] ]

[ "竞赛->知识点->数与式->二次根式->二次根式的性质与运算" ]

[ "$$\\sqrt{a{{b}^{7}}{{c}^{3}}}={{b}^{3}}c\\sqrt{abc}$$，$$\\sqrt{\\frac{{{c}^{3}}}{ab}}=\\frac{c}{ab}\\sqrt{abc}$$，它们是同类二次根式． 故选$$\\text{D}$$． " ]

[ "2009年第20届希望杯初一竞赛第1试第5题4分" ]

[ [ { "aoVal": "A", "content": "锐角三角形 " } ], [ { "aoVal": "B", "content": "钝角三角形 " } ], [ { "aoVal": "C", "content": "有一个角是$$30{}^{}\\circ $$的直角三角形 " } ], [ { "aoVal": "D", "content": "等腰直角三角形 " } ] ]

[ "课内体系->能力->分析和解决问题能力", "课内体系->知识点->三角形->三角形及多边形->三角形的基础->三角形的分类", "课内体系->知识点->三角形->三角形及多边形->与三角形有关的角->三角形内角和的应用" ]

[ "由题意得：$$\\begin{cases}\\angle A=\\dfrac{3}{2}\\angle B \\angle C=\\angle A+\\angle B+30{}^{}\\circ \\angle A+\\angle B+\\angle C=180{}^{}\\circ \\end{cases}$$， 解得$$\\angle A=45{}^{}\\circ $$，$$\\angle B=30{}^{}\\circ $$，$$\\angle C=105{}^{}\\circ $$． 故选$$\\text{B}$$． " ]

[ "2001年第12届希望杯初二竞赛第1试第4题" ]

[ [ { "aoVal": "A", "content": "一切实数 " } ], [ { "aoVal": "B", "content": "一切正实数 " } ], [ { "aoVal": "C", "content": "一切大于或等于$$5$$的实数 " } ], [ { "aoVal": "D", "content": "一切大于或等于$$2$$的实数 " } ] ]

[ "竞赛->知识点->函数->二次函数->二次函数的最大值与最小值" ]

[ "$$y={{x}^{4}}-4{{x}^{3}}+8{{x}^{2}}-8x+5$$ $$={{x}^{4}}+4{{x}^{2}}+4-4{{x}^{3}}+4{{x}^{2}}-8x+1$$ $$={{({{x}^{2}}+2)}^{2}}-4x({{x}^{2}}+2)+{{(2x)}^{2}}+1$$ $$={{[({{x}^{2}}+2)-2x]}^{2}}+1$$ $$={{[{{(x-1)}^{2}}+1]}^{2}}+1$$. 因为$${{(x-1)}^{2}}\\geqslant 0$$，所以$${{(x-1)}^{2}}+1\\geqslant 1$$. 所以 当$$x=1$$时，$$y$$取得最小值$$2$$， 即$$y$$的取值范围是一切大于或等于$$2$$的实数．故选$$\\text{D}$$. " ]

[ "1996年第7届全国希望杯初一竞赛初赛第4题" ]

[ [ { "aoVal": "A", "content": "是同解方程 " } ], [ { "aoVal": "B", "content": "不是同解方程 " } ], [ { "aoVal": "C", "content": "是同一个方程 " } ], [ { "aoVal": "D", "content": "可能不是同解方程 " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->一元一次方程->解一元一次方程->方程解的关系" ]

[ "当另一个方程的解也都满足第一个方程时，这两个方程才是同解方程，因此排除$$\\text{B}$$． 但另一个方程的解不都满足第一个方程时，它们不是同解方程，所以排除$$\\text{A}$$、$$\\text{C}$$， 因此选$$\\text{D}$$． " ]

[ "2011年第16届华杯赛初一竞赛初赛第1题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{7}{14}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{9}{14}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{2}{9}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{4}{9}$$ " } ] ]

[ "竞赛->知识点->方程与不等式->方程应用" ]

[ "由已知条件可设船顺水航行的速度是$$7v$$，逆水航行的速度是$$2v$$，设两港之间的距离为$$s$$，则船往返一次的平均速度为$$\\frac{2s}{\\frac{s}{7v}+\\frac{s}{2v}}=\\frac{28}{9}v$$，从而平均速度与顺水速度之比为$$\\frac{28}{9}v:7v=\\frac{4}{9}$$． 故选$$\\text{D}$$． " ]

[ "2008年第19届希望杯初一竞赛第2试第4题4分" ]

[ [ { "aoVal": "A", "content": "①、② " } ], [ { "aoVal": "B", "content": "①、③ " } ], [ { "aoVal": "C", "content": "②、③ " } ], [ { "aoVal": "D", "content": "③、④ " } ] ]

[ "竞赛->知识点->数论->整除->整除的概念与基本性质" ]

[ "由题意，得$$a=2$$，$$b=3$$，$$c=5$$， $${{(a+b)}^{2}}=25$$，$$c=5$$，①不正确； $${{a}^{2}}+{{b}^{2}}=13$$，$$c=5$$，②正确； $${{(b+c)}^{2}}=64$$，$$a=2$$，③不正确； $${{a}^{2}}+{{c}^{2}}=29$$，$$b=3$$，④正确； 不正确的判断是①、③． 故选$$\\text{B}$$． " ]

[ "1995年第6届全国希望杯初一竞赛初赛第5题" ]

[ [ { "aoVal": "A", "content": "$$34.49$$ " } ], [ { "aoVal": "B", "content": "$$34.51$$ " } ], [ { "aoVal": "C", "content": "$$34.99$$ " } ], [ { "aoVal": "D", "content": "$$35.01$$ " } ] ]

[ "课内体系->知识点->数->有理数->近似数与有效数字->精确数位" ]

[ "由于$$34.51$$，$$34.99$$，$$35.01$$四舍五入的近似值都可能是$$35$$， 而只有$$34.49$$不可能是真值，选$$\\text{A}$$． " ]

[ "2012年竞赛第5题3分", "初二其它" ]

[ [ { "aoVal": "A", "content": "$$2012$$ " } ], [ { "aoVal": "B", "content": "$$101$$ " } ], [ { "aoVal": "C", "content": "$$100$$ " } ], [ { "aoVal": "D", "content": "$$99$$ " } ] ]

[ "课内体系->知识点->式->因式分解->因式分解的应用", "课内体系->能力->运算能力" ]

[ "由$$a+b+ab+1=(a+1)(b+1)$$得，每次操作前和操作后，黑板上的每个数加$$1$$后的乘积不变， 设经过$$99$$次操作后黑板上剩下的数为$$x$$， 则$$x+1=(1+1)( \\frac{1}{2}+1)(\\frac{1}{3}+1)\\cdots (\\frac{1}{100}+1)$$， 解得$$x=100$$． " ]

[ "2016年全国华杯赛初一竞赛" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->式->整式的乘除->乘法公式->完全平方公式的计算-不含分式", "课内体系->知识点->式->分式->分式的运算->分式加减混合运算" ]

[ "∵$$\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=5$$，$$xyz=1$$， ∴$$\\frac{yz+xz+xy}{xyz}=5$$，即$$xy+yz+xz=5$$． ∵$$x+y+z=5$$， ∴$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{\\left( x+y+z \\right)}^{2}}-2\\left( xy+yz+xz \\right)=15$$． " ]

[ "2006年第17届希望杯初二竞赛第1试第9题" ]

[ [ { "aoVal": "A", "content": "$$2\\sqrt{3}$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$\\pm 2\\sqrt{3}$$ " } ], [ { "aoVal": "D", "content": "$$4$$或$$\\pm 2\\sqrt{3}$$ " } ] ]

[ "课内体系->知识点->式->整式的乘除->整式的乘除运算->单项式乘单项式", "课内体系->知识点->综合与实践->规律探究与程序框图->定义新运算", "课内体系->能力->运算能力", "课内体系->思想->分类讨论思想" ]

[ "当$$3\\geqslant m$$时，有 $$3*m={{3}^{2}}\\cdot m=9m=36$$， 解得$$m=4$$，与$$3\\geqslant m$$予盾，舍去． 当$$3\\textless{}m$$时，有$$3*m=3{{m}^{2}}=36$$， 解得$$m=\\pm 2\\sqrt{3}$$，因为$$3\\textless{}m$$，因此舍去$$m=-2\\sqrt{3}$$． 所以选$$\\text{A}$$． " ]

[ "2019~2020学年10月江苏无锡梁溪区东林中学初一上学期月考第10题3分", "1995年第6届全国希望杯初一竞赛复赛第1题", "2017~2018学年10月江苏扬州高邮市高邮市南海中学初一上学期月考第6题3分" ]

[ [ { "aoVal": "A", "content": "$${{x}^{3}}y\\textgreater0$$ " } ], [ { "aoVal": "B", "content": "$$x+\\left\\textbar{} y \\right\\textbar\\textless{}0$$ " } ], [ { "aoVal": "C", "content": "$$\\left\\textbar{} x \\right\\textbar+y\\textgreater0$$ " } ], [ { "aoVal": "D", "content": "$$x-{{y}^{2}}\\textless{}0$$ " } ], [ { "aoVal": "E", "content": "Both A and C " } ] ]

[ "课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质" ]

[ "∵$$y\\textgreater0$$，若$$x\\geqslant 0$$则$$x+y\\textgreater0$$，与$$x+y\\textless{}0$$矛盾． 所以由$$y\\textgreater0$$，$$x+y\\textless{}0$$必有$$x\\textless{}0$$． 因此，$${{x}^{3}}\\textless{}0$$，$${{x}^{3}}y\\textless{}0$$，即$$\\text{A}$$是错误的． 事实上，$$y\\textgreater0$$，$$x+y\\textless{}0$$，即$$x+\\left\\textbar{} y \\right\\textbar\\textless{}0$$，$$\\text{B}$$成立． $$\\left\\textbar{} x \\right\\textbar+y\\textgreater0$$，$$\\text{C}$$成立．$$x\\textless{}0$$，$${{y}^{2}}\\textgreater0$$，$$x-{{y}^{2}}\\textless{}0$$，$$\\text{D}$$成立．因此，选$$\\text{A}$$． " ]

[ "2000年第17届全国初中数学联赛竞赛第1题7分" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$\\sqrt{5}$$ " } ], [ { "aoVal": "C", "content": "$$2\\sqrt{5}$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->式->二次根式->二次根式的运算->二次根式的四则混合运算", "课内体系->知识点->式->二次根式->二次根式的运算->多重二次根式" ]

[ "$$\\sqrt{14+6\\sqrt{5}}-\\sqrt{14-6\\sqrt{5}}=\\sqrt{14+2\\sqrt{45}}-\\sqrt{14-2\\sqrt{45}}=\\sqrt{{{(\\sqrt{9}+\\sqrt{5})}^{2}}}-\\sqrt{{{(\\sqrt{9}-\\sqrt{5})}^{2}}}$$， $$=\\sqrt{9}+\\sqrt{5}-\\sqrt{9}+\\sqrt{5}=2\\sqrt{5}$$． " ]

[ "2001年第12届希望杯初二竞赛第1试第2题" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1+\\sqrt{2}$$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "不存在的 " } ] ]

[ "课内体系->知识点->式->二次根式->二次根式的基础->二次根式有意义的条件", "课内体系->知识点->式->二次根式->二次根式的运算->二次根式的加减以及混合运算", "课内体系->知识点->式->二次根式->二次根式化简求值" ]

[ "要使代数式$$\\sqrt{x}+\\sqrt{x-1}+\\sqrt{x-2}$$有意义，需满足条件： 当$$\\begin{cases}x\\geqslant 0 x-1\\geqslant 0 x-2\\geqslant 0 \\end{cases}$$，所以$$x\\geqslant 2$$， 当$$x\\geqslant 2$$时， 即$$\\sqrt{x}+\\sqrt{x-1}+\\sqrt{x-2}\\geqslant \\sqrt{2}+\\sqrt{2-1}+\\sqrt{2-2}=\\sqrt{2}+1$$， 即$$x=2$$时，原代数式取得最小值$$\\sqrt{2}+1$$． 故选$$\\text{B}$$． " ]

[ "2014年第25届全国希望杯初二竞赛初赛第10题4分" ]

[ [ { "aoVal": "A", "content": "$$1152$$种 " } ], [ { "aoVal": "B", "content": "$$576$$种 " } ], [ { "aoVal": "C", "content": "$$288$$种 " } ], [ { "aoVal": "D", "content": "$$144$$种 " } ] ]

[ "课内体系->能力->分析和解决问题能力", "竞赛->知识点->组合->排列与组合" ]

[ "因为去掉$$8$$以后，余下$$7$$个数的和是$$1+2+3+4+5+6+7=28$$， 所以$$8$$的两边各数的和分别是$$14$$． 因为$$7+6=13$$， 所以$$14$$至少是由$$3$$个数相加而成的， 所以$$8$$的两边分别有$$3$$个数和$$4$$个数． 因为$$(7,6,1)$$，$$(7,5,2)$$，$$(7,4,3)$$，$$(6,5,3)$$这四组数的和都是$$14$$， 每组中的数的排列方法有$$3\\times 2\\times 1=6$$种， 与其对应的另外$$4$$个数（$$8$$除外）有$$4\\times 3\\times 2\\times 1=24$$种排列方法， 当排成横行时，这$$4$$组数有排在$$8$$的左边和右边两种排列方法， 因此，每组数的排列方法有$$2\\times 6\\times 24=288$$（种）． 所以，满足题意的排列方法有$$4\\times 288=1152$$（种）． " ]

[ "初二上学期其它", "2002年竞赛第1题5分", "2015年山东青岛黄岛区山东省青岛第九中学自主招生", "2016~2017学年江西景德镇初二下学期期中景德镇一中1班第1题4分" ]

[ [ { "aoVal": "A", "content": "$$\\sqrt{3}$$ " } ], [ { "aoVal": "B", "content": "$$\\sqrt{6}$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ] ]

[ "知识标签->题型->式->分式->分式化简求值->题型：分式条件化简求值", "知识标签->学习能力->运算能力", "知识标签->知识点->式->分式->分式的运算->分式的混合运算" ]

[ "$$a+b=-\\sqrt{6ab}$$，$$a-b=-\\sqrt{2ab}$$． ∵$${{a}^{2}}+{{b}^{2}}=4ab$$， ∴$${{a}^{2}}+{{b}^{2}}+2ab={{\\left( a+b \\right)}^{2}}=6ab$$① ∴$${{a}^{2}}+{{b}^{2}}-2ab={{\\left( a-b \\right)}^{2}}=2ab$$② $$\\frac{①}{②}$$，得~~$$\\frac{{{\\left( a+b \\right)}^{2}}}{{{\\left( a-b \\right)}^{2}}}=\\frac{6ab}{2ab}$$ ∵$$a\\textless{}b\\textless{}0$$， ∴$$ab\\textgreater0$$，$$a+b\\textless{}0$$，$$a-b\\textless{}0$$， ∴$$\\frac{{{\\left( a+b \\right)}^{2}}}{{{\\left( a-b \\right)}^{2}}}={{\\left( \\frac{a+b}{a-b} \\right)}^{2}}=3$$， ∴$$\\frac{a+b}{a-b}=\\sqrt{3}$$． 故选：$$\\text{A}$$． " ]

[ "2016年山东青岛竞赛二中杯", "初二其它" ]

[ [ { "aoVal": "A", "content": "$$2012$$ " } ], [ { "aoVal": "B", "content": "$$101$$ " } ], [ { "aoVal": "C", "content": "$$100$$ " } ], [ { "aoVal": "D", "content": "$$99$$ " } ] ]

[ "课内体系->知识点->式->因式分解->其他方法", "课内体系->知识点->式->因式分解->因式分解的应用" ]

[ "由$$a+b+ab+1=(a+1)(b+1)$$得，每次操作前和操作后，黑板上的每个数加$$1$$后的乘积不变， 设经过$$99$$次操作后黑板上剩下的数为$$x$$， 则$$x+1=(1+1)( \\frac{1}{2}+1)(\\frac{1}{3}+1)\\cdots (\\frac{1}{100}+1)$$， 解得$$x=100$$． " ]

[ "2002年第13届希望杯初一竞赛第2试第2题" ]

[ [ { "aoVal": "A", "content": "1个 " } ], [ { "aoVal": "B", "content": "2个 " } ], [ { "aoVal": "C", "content": "3个 " } ], [ { "aoVal": "D", "content": "4个 " } ] ]

[ "课内体系->知识点->几何图形初步->相交线与平行线->相交线->对顶角的定义与性质", "课内体系->知识点->几何图形初步->相交线与平行线->相交线->计算对顶角、邻补角度数" ]

[ "②两个角相等，不一定为对顶角； ③两个角不是对项角，也可以相等． " ]

[ "1996年第7届全国希望杯初一竞赛初赛第7题" ]

[ [ { "aoVal": "A", "content": "$$254$$厘米 " } ], [ { "aoVal": "B", "content": "$$150$$厘米 " } ], [ { "aoVal": "C", "content": "$$127$$厘米 " } ], [ { "aoVal": "D", "content": "$$871$$厘米 " } ] ]

[ "课内体系->知识点->几何图形初步->直线、射线、线段->直线、射线、线段的运算->线段和差的计算问题->线段和差-需要分类讨论" ]

[ "$$PQ=AQ+PB-AB=1200+1050-1996=254$$（厘米），选$$\\text{A}$$． " ]

[ "2013年第18届华杯赛初一竞赛初赛第2题" ]

[ [ { "aoVal": "A", "content": "$$-\\frac{2}{3}$$ " } ], [ { "aoVal": "B", "content": "$$-\\frac{3}{4}$$ " } ], [ { "aoVal": "C", "content": "$$-1$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{2}{3}$$ " } ], [ { "aoVal": "E", "content": "$$-\\frac{1}{3}$$ " } ] ]

[ "竞赛->知识点->数与式->整式->整式的乘除运算", "课内体系->知识点->数->有理数->分数" ]

[ "由条件可知，$$a=(-6)\\times (-4)=24$$，$$b=(-6)\\times 6=-36$$，所以 $$\\frac{a}{b}=-\\frac{24}{36}=-\\frac{2}{3}$$． 故选$$\\text{A}$$． " ]

[ "1998年第9届希望杯初二竞赛第1试第10题" ]

[ [ { "aoVal": "A", "content": "$${{a}_{1}}+{{a}_{2}}\\leqslant {{b}_{1}}+{{b}_{2}}$$ " } ], [ { "aoVal": "B", "content": "$${{a}_{1}}+{{a}_{2}}\\geqslant {{b}_{1}}+{{b}_{2}}$$ " } ], [ { "aoVal": "C", "content": "$${{a}_{1}}+{{a}_{2}}={{b}_{1}}+{{b}_{2}}$$ " } ], [ { "aoVal": "D", "content": "无法确定的 " } ] ]

[ "课内体系->能力->分析和解决问题能力", "课内体系->知识点->方程与不等式->不等式（组）->不等式->不等式的性质" ]

[ "因为$${{a}_{1}}$$，$${{a}_{2}}$$，$${{b}_{1}}$$，$${{b}_{2}}$$均为正数，且$${{a}_{1}}{{a}_{2}}\\leqslant {{b}_{1}}{{b}_{2}}$$． 所以$${{b}_{2}}\\geqslant \\frac{{{a}_{1}}{{a}_{2}}}{{{b}_{1}}}$$， 所以$${{b}_{1}}+{{b}_{2}}\\geqslant {{b}_{1}}+\\frac{{{a}_{1}}{{a}_{2}}}{{{b}_{1}}}$$， 所以$$({{b}_{1}}+{{b}_{2}})-({{a}_{1}}+{{a}_{2}})\\geqslant {{b}_{1}}+\\frac{{{a}_{1}}{{a}_{2}}}{{{b}_{1}}}-{{a}_{1}}-{{a}_{2}}$$ $$=({{b}_{1}}-{{a}_{1}})+\\frac{{{a}_{2}}({{a}_{1}}-{{b}_{1}})}{{{b}_{1}}}=({{b}_{1}}-{{a}_{1}})\\left( 1-\\frac{{{a}_{2}}}{{{b}_{1}}} \\right)$$． 因为$${{a}_{1}}\\leqslant {{b}_{1}}$$，$${{a}_{2}}\\leqslant {{a}_{1}}$$， 所以$$0\\textless{}{{a}_{2}}\\leqslant {{b}_{1}}$$，$$0\\textless{}\\frac{{{a}_{2}}}{{{b}_{1}}}\\leqslant 1$$． 所以$${{b}_{1}}-{{a}_{1}}\\geqslant 0$$，$$\\left( 1-\\frac{{{a}_{2}}}{{{b}_{1}}} \\right)\\geqslant 0$$． 所以$${{b}_{1}}+{{b}_{2}}\\geqslant {{a}_{1}}+{{a}_{2}}$$． 又当$${{a}_{1}}$$，$${{a}_{2}}$$，$${{b}_{1}}$$，$${{b}_{2}}$$均相等时，等号成立． 故选$$\\text{A}$$． " ]

[ "1999年竞赛（全国初中数学竞赛）第2题5分" ]

[ [ { "aoVal": "A", "content": "$$60$$元 " } ], [ { "aoVal": "B", "content": "$$66$$元 " } ], [ { "aoVal": "C", "content": "$$70$$元 " } ], [ { "aoVal": "D", "content": "$$75$$元 " } ] ]

[ "课内体系->能力->运算能力", "课内体系->能力->推理论证能力", "课内体系->能力->分析和解决问题能力", "课内体系->知识点->方程与不等式->一元一次方程->一元一次方程与实际问题->一元一次方程的梯度计价问题" ]

[ "由于$$0.88\\textgreater0.8$$，所以这户人家用的煤气超过$$60$$立方米．设用了$$x$$立方米，则 $$60\\cdot 0.8+1.2\\left( x-60 \\right)=0.88x$$， 解得$$x=75$$， $$75\\times 0.88=66$$（元）． 故选：$$\\text{B}$$． " ]

[ "初二下学期单元测试《二次根式》二次根式的运算第43题", "2002年第19届全国初中数学联赛竞赛第1题7分" ]

[ [ { "aoVal": "A", "content": "$$a\\textless{}b\\textless{}c$$ " } ], [ { "aoVal": "B", "content": "$$a\\textless{}c\\textless{}b$$ " } ], [ { "aoVal": "C", "content": "$$b\\textless{}a\\textless{}c$$ " } ], [ { "aoVal": "D", "content": "$$b\\textless{}c\\textless{}a$$ " } ] ]

[ "课内体系->知识点->式->二次根式->二次根式的运算->二次根式比较大小" ]

[ "∵$$a-b$$ $$=\\sqrt{2}-1-\\left( 2\\sqrt{2}-\\sqrt{6} \\right)$$ $$=\\sqrt{6}-\\left( 1+\\sqrt{2} \\right)$$ $$\\approx 2.449-2.414\\textgreater0$$， ∴$$a\\textgreater b$$； ∵$$a-c$$ $$=\\sqrt{2}-1-\\left( \\sqrt{6}-2 \\right)$$ $$=\\sqrt{2}+1-\\sqrt{6}$$ $$\\approx 2.414-2.449\\textless{}0$$， ∴$$a\\textless{}c$$； 于是$$b\\textless{}a\\textless{}c$$． " ]

[ "2010年第21届全国希望杯初一竞赛初赛第10题4分", "2019~2020学年浙江宁波镇海区宁波市镇海蛟川书院初一上学期期中第8题3分" ]

[ [ { "aoVal": "A", "content": "存在负整数 " } ], [ { "aoVal": "B", "content": "存在正整数 " } ], [ { "aoVal": "C", "content": "存在负分数 " } ], [ { "aoVal": "D", "content": "不存在正分数 " } ] ]

[ "课内体系->知识点->数->有理数->相反数->相反数的性质" ]

[ "由可知$$a\\ne 0$$且$$b\\ne 0$$． 再由$$a+b=0$$，可知$$a=-b$$． 不妨设$$a\\textgreater b$$，则$$a\\textgreater0\\textgreater b$$， 则在$$a$$和$$b$$之间一定存在负分数． " ]

[ "1992年第3届全国希望杯初一竞赛复赛第7题" ]

[ [ { "aoVal": "A", "content": "$$x$$是奇数，$$y$$是偶数 " } ], [ { "aoVal": "B", "content": "$$x$$是偶数，$$y$$是奇数 " } ], [ { "aoVal": "C", "content": "$$x$$是偶数，$$y$$是偶数 " } ], [ { "aoVal": "D", "content": "$$x$$是奇数，$$y$$是奇数 " } ] ]

[ "课内体系->思想->方程思想", "课内体系->思想->整体思想", "课内体系->知识点->方程与不等式->二元一次方程（组）->解二元一次方程组->二元一次方程组的定义" ]

[ "∵$$p$$为偶数，$$q$$为奇数，其解$$x$$，$$y$$又是整数， 由$$x-1992y=p$$可知$$x$$为偶数，由$$1993x+3y=q$$可知$$y$$是奇数，选$$\\text{B}$$． " ]

[ "2000年第11届希望杯初一竞赛第6题" ]

[ [ { "aoVal": "A", "content": "$$25 \\%$$ " } ], [ { "aoVal": "B", "content": "$$40 \\%$$ " } ], [ { "aoVal": "C", "content": "$$50 \\%$$ " } ], [ { "aoVal": "D", "content": "$$66.7 \\%$$ " } ] ]

[ "课内体系->能力->分析和解决问题能力", "课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->一元一次方程->一元一次方程与实际问题->一元一次方程的经济问题->一元一次方程的经济问题-打折" ]

[ "设标价为$$x$$，进价为$$a$$，由题意得$$0.8x-a=0.2a$$解得$$x=1.5a$$，按原价出售可获利$$\\frac{1.5a-a}{a}=0.5$$． " ]

[ "1997年第8届全国希望杯初一竞赛初赛第10题" ]

[ [ { "aoVal": "A", "content": "小于或等于$$3$$的有理数 " } ], [ { "aoVal": "B", "content": "小于$$3$$的有理数 " } ], [ { "aoVal": "C", "content": "小于或等于$$-3$$的有理数 " } ], [ { "aoVal": "D", "content": "小于$$-3$$的有理数 " } ] ]

[ "课内体系->知识点->数->有理数->绝对值->绝对值的非负性", "课内体系->知识点->数->有理数->绝对值->绝对值的代数意义", "课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式->解一元一次不等式" ]

[ "$$\\left\\textbar{} b \\right\\textbar\\textless{}3$$就是$$-3\\textless{}b\\textless{}3$$，只有当$$a\\leqslant -3$$时，$$a\\textless{}b$$恒成立，选$$\\text{C}$$． " ]

[ "2018年第29届希望杯初一竞赛初赛第7题4分" ]

[ [ { "aoVal": "A", "content": "$$267$$ " } ], [ { "aoVal": "B", "content": "$$268$$ " } ], [ { "aoVal": "C", "content": "$$270$$ " } ], [ { "aoVal": "D", "content": "$$272$$ " } ] ]

[ "课内体系->能力->运算能力", "课内体系->能力->分析和解决问题能力", "课内体系->知识点->数->有理数->有理数与实际问题->有理数乘除法与实际问题" ]

[ "在$$\\text{A}$$商店每购$$5$$本赠一本，$$5\\div6 \\approx0.83$$，低于八五折， 所以可以在$$\\text{A}$$商店购买$$24$$本，在$$\\text{B}$$商店购买$$8$$本； 或者在$$\\text{A}$$商店购买直接购买$$32$$本． 在$$\\text{A}$$商店购买$$24$$本，在$$\\text{B}$$商店购买$$8$$本， 花费$$20\\times10+8\\times10\\times0.85=268$$（元）； 在$$\\text{A}$$商店购买直接购买$$32$$本，花费$$25\\times10+2\\times10=270$$（元）． 综上，小明需要购买$$32$$本笔记本，则他至少花$$268$$元． 故选$$\\text{B}$$． " ]

[ "2000年第11届希望杯初二竞赛第1试第8题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{a}$$，$$\\frac{1}{b}$$，$$\\frac{1}{c}$$ " } ], [ { "aoVal": "B", "content": "$$a^{2}$$，$$ b^{2}$$，$$c^{2}$$ " } ], [ { "aoVal": "C", "content": "$$\\sqrt{a}$$，$$\\sqrt{b}$$，$$ \\sqrt{c}$$ " } ], [ { "aoVal": "D", "content": "$$\\textbar a-b\\textbar$$，$$\\textbar b-c\\textbar$$，$$\\textbar c-a\\textbar$$ " } ] ]

[ "竞赛->知识点->三角形->三角形基础" ]

[ "如$$2$$，$$4$$，$$5$$为边可以组成三角形，而$$\\frac{1}{2}$$，$$ \\frac{1}{4}$$，$$ \\frac{1}{5}$$不能构成三角形，排除$$\\text{A}$$． 如$$3$$，$$4$$，$$5$$为边可以组成三角形，而$$3^{2}$$，$$ 4^{2}$$，$$ 5^{2}$$不能构成三角形，排除$$\\text{B}$$． 如$$2$$，$$3$$，$$4$$为边可以组成三角形，而$$\\textbar2-3\\textbar$$，$$\\textbar3-4\\textbar$$，$$\\textbar4-2\\textbar$$不能构成三角形，排除$$\\text{D}$$． 不妨设$$a\\textgreater b\\textgreater c\\textgreater0$$，满足$$a\\textless b+c$$， 所以$$(\\sqrt{b}+\\sqrt{c})^{2}=b+2 \\sqrt{b c}+c\\textgreater a$$， 所以$$\\sqrt{b}+\\sqrt{c}\\textgreater\\sqrt{a}$$， 所以$$\\sqrt{a}$$，$$ \\sqrt{b}$$，$$ \\sqrt{c}$$可以构成三角形． 故选$$\\text{C}$$． " ]

[ "2016年第27届全国希望杯初二竞赛初赛第5题4分" ]

[ [ { "aoVal": "A", "content": "$$-1\\textless{}m\\textless{}1$$ " } ], [ { "aoVal": "B", "content": "$$-1\\leqslant m\\textless{}1$$ " } ], [ { "aoVal": "C", "content": "$$-1\\textless{}m\\leqslant 1$$ " } ], [ { "aoVal": "D", "content": "$$-1\\leqslant m\\leqslant 1$$ " } ] ]

[ "课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式->解一元一次不等式", "课内体系->知识点->方程与不等式->不等式（组）->含参不等式->由不等式（组）的整数解情况求参数范围" ]

[ "解$$2x-1\\textgreater3(x-2)$$，得$$x\\textless{}5$$； 解$$2x+m-1\\textgreater0$$，得$$x\\textgreater\\frac{1-m}{2}$$． ∵不等式组只有四个整数解， ∴$$0\\leqslant \\frac{1-m}{2}\\textless{}1$$， 解得$$-1\\textless{}m\\leqslant 1$$． " ]

[ "初二下学期其它", "1996年第7届希望杯初二竞赛第4题" ]

[ [ { "aoVal": "A", "content": "$$a\\textgreater b\\textgreater c\\textgreater d$$ " } ], [ { "aoVal": "B", "content": "$$c\\textgreater a\\textgreater d\\textgreater b$$ " } ], [ { "aoVal": "C", "content": "$$a\\textgreater d\\textgreater c\\textgreater b$$ " } ], [ { "aoVal": "D", "content": "$$a\\textgreater c\\textgreater d\\textgreater b$$ " } ] ]

[ "课内体系->知识点->数->有理数->有理数比较大小->有理数比较大小-利用有理数正负性", "课内体系->知识点->数->有理数->有理数基础运算->有理数除法->有理数除法运算", "课内体系->能力->运算能力" ]

[ "$$a=10001+\\dfrac{10000}{1995}$$，$$b=10001-\\dfrac{10000}{1996}$$，$$c=10001+\\dfrac{1}{1995}$$，$$d=10001-\\dfrac{1}{1996}$$． 所以$$a\\textgreater c\\textgreater d\\textgreater b$$． " ]

[ "1994年第5届希望杯初二竞赛第7题" ]

[ [ { "aoVal": "A", "content": "$$182$$ " } ], [ { "aoVal": "B", "content": "$$180$$ " } ], [ { "aoVal": "C", "content": "$$32$$ " } ], [ { "aoVal": "D", "content": "$$30$$ " } ] ]

[ "竞赛->知识点->三角形->特殊三角形->直角三角形" ]

[ "设另一条直角边的长度为$$x$$，斜边的长度为$$z$$，则$${{z}^{2}}-{{x}^{2}}={{13}^{2}}$$， 由于$$x$$，$$z$$均为正整数，且$$z\\textgreater x$$， 于是由$$\\left( z+x \\right)\\left( z-x \\right)=169\\times 1$$． 可得$$\\begin{cases}z+x=169 z-x=1 \\end{cases}$$． 所以三角形的周长为$$169+13=182$$． 故选$$\\text{A}$$． " ]

[ "1998年第9届希望杯初二竞赛第1试第1题" ]

[ [ { "aoVal": "A", "content": "$$(x+2y-3z)(x-2y-3z)$$ " } ], [ { "aoVal": "B", "content": "$$(x-2y-3z)(x-2y+3z)$$ " } ], [ { "aoVal": "C", "content": "$$(x+2y+3z)(x+2y-3z)$$ " } ], [ { "aoVal": "D", "content": "$$(x+2y+3z)(x-2y-3z)$$ " } ] ]

[ "竞赛->知识点->数与式->因式分解->因式分解：公式法" ]

[ "$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde{{x}^{2}}-4{{y}^{2}}-9{{z}^{2}}-12yz$$ $$={{x}^{2}}-(4{{y}^{2}}+12yz+9{{z}^{2}})$$ $$={{x}^{2}}-{{(2y+3z)}^{2}}$$ $$=[x+(2y+3z)][x-(2y+3z)]$$ $$=(x+2y+3z)(x-2y-3z)$$． 故选$$\\text{D}$$． " ]

[ "1991年第2届希望杯初二竞赛第2题" ]

[ [ { "aoVal": "A", "content": "$$ts$$ " } ], [ { "aoVal": "B", "content": "$$s-ts$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{ts}{1+s}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{s}{1+t}$$ " } ] ]

[ "竞赛->知识点->数与式->数的运算->有理数运算问题" ]

[ "设$$m=3$$，$$n=1$$，则$$t=3$$，$$s=4$$，只有$$\\frac{s}{1+t}=\\frac{4}{1+3}=1=n$$． 故选$$\\text{D}$$． " ]

[ "2014年全美数学竞赛（AMC）竞赛" ]

[ [ { "aoVal": "A", "content": "all $$4$$ are boys 4个都是男孩 " } ], [ { "aoVal": "B", "content": "all $$4$$ are girls 四个都是女孩 " } ], [ { "aoVal": "C", "content": "$$2$$ are girls and $$2$$ are boys 2个女孩和$$2$$个男孩 " } ], [ { "aoVal": "D", "content": "$$3$$ are of one gender and $$1$$ is of the other gender 3个是同一种性别，$$1$$个是另一种性别 " } ], [ { "aoVal": "E", "content": "all of these outcomes are equally likely 所有这些结果都有同样的可能性 " } ] ]

[ "美国AMC8->Knowledge Point->Counting, Probability and Statistics->Statistical Distribution", "课内体系->知识点->统计与概率->概率->可能性的大小" ]

[ "城市医院昨日诞下四名婴儿．假设每个婴儿各可能有一半几率为男孩或女孩．则下列哪个事件最有可能？ A．$$4$$个男婴 B．$$4$$个女婴 C．$$2$$男婴$$2$$女婴 D．$$3$$个为一种性别，另一个为异性 E．以上事件皆等可能 我们先把例子分析一下．$$A$$发生的概率为$${{\\left( \\frac{1}{2} \\right)}^{4}}=\\frac{1}{16}$$，$$B$$发生的概率为$${{\\left( \\frac{1}{2} \\right)}^{4}}=\\frac{1}{16}$$．$$C$$发生的概率为$$\\left( \\begin{matrix}4 2 \\end{matrix} \\right)\\cdot {{\\left( \\frac{1}{2} \\right)}^{4}}=\\frac{3}{8}$$．因为我们需要从$$4$$个孩子中选择$$2$$个成为女孩． $$D$$方面，有两种可能的情况，即$$3$$个女孩和$$1$$个男孩或$$3$$个男孩和$$1$$个女孩．第一种情况的概率就$$\\left( \\begin{matrix}4 1 \\end{matrix} \\right)\\cdot {{\\left( \\frac{1}{2} \\right)}^{4}}=\\frac{1}{4}$$．因为我们需要从四个孩子中选择一个成为男孩．但是，第二种情况也有同样的可能性，因为我们在$$4$$个孩子中选择了$$1$$个是女孩，所以总的可能性是$$\\frac{1}{4}\\cdot 2=\\frac{1}{2}$$． 所以在四个分数中，$$D$$就最大的．所以我们的答案是$$3$$ are of one gender and $$1$$ is of the other gender． 故选$$\\text{D}$$． 可能性列在帕斯卡三角形的第四行，最左边的$$1$$是所有男孩的可能性，最右边的$$1$$是所有女孩的可能性．由于 pascal 的三角形第四行为$$1$$，$$4$$，$$6$$，$$4$$，$$1$$，$$6$$，均为每个性别的两个孩子的可能性，因此共有$$8$$个可能性，其中一个性别的三个孩子和另一个性别的一个孩子．由于总共有$$2^{4}=16$$个儿童性别的可能性． 故选$$\\text{D}$$． We\\textquotesingle ll just start by breaking cases down. The probability of $$A$$ occurring is $${{\\left( \\frac{1}{2} \\right)}^{4}}=\\frac{1}{16}$$. The probability of $$B$$ occurring is $${{\\left( \\frac{1}{2} \\right)}^{4}}=\\frac{1}{16}$$. The probability of $$C$$ occurring is $$\\left( \\begin{matrix}4 2 \\end{matrix} \\right)\\cdot {{\\left( \\frac{1}{2} \\right)}^{4}}=\\frac{3}{8}$$, because we need to choose $$2$$ of the $$4$$ children to be girls. For $$D$$, there are two possible cases, $$3$$ girls and $$1$$ boy or $$3$$ boys and $$1$$ girl. The probability of the first case is $$\\left( \\frac{4}{1} \\right)\\cdot {{\\left( \\frac{1}{2} \\right)}^{4}}=\\frac{1}{4}$$ because we need to choose $$1$$ of the $$4$$ children to be a boy. However, the second case has the same probability because we are choosing $$1$$ of the $$4$$ children to be a girl, so the total probability is $$\\frac{1}{4}\\cdot 2=\\frac{1}{2}$$. So out of the four fractions, $$D$$ is the largest. So our answer is $$\\left( \\text{D} \\right)$$ $$3$$ of one gender and $$1$$ of the other. The possibilities are listed out in the fourth row of Pascal\\textquotesingle s triangle, with the leftmost $$1$$ being the possibility of all boys and the rightmost $$1$$ being the possibility of all girls. Since the fourth row of Pascal\\textquotesingle s Triangle goes $$1$$, $$4$$, $$6$$, $$4$$, $$1$$ and $$6$$ are all the possibilities of two children from each gender, there are a total of $$8$$ possibilities of three children from one gender and one from the other. Since there are a total of $${{2}^{4}}=16$$ total possibilities for the gender of the children, $$\\text{D}$$ has the highest probability. " ]

[ "2018~2019学年9月湖南长沙雨花区中雅培粹学校初一上学期月考第12题3分", "1999年第10届希望杯初一竞赛第10题" ]

[ [ { "aoVal": "A", "content": "$$({{a}_{1}}+1)({{a}_{2}}+2)\\cdots ({{a}_{n}}+n)$$是正整数 " } ], [ { "aoVal": "B", "content": "$$({{a}_{1}}-1)({{a}_{2}}-2)\\cdots ({{a}_{n}}-n)$$是正整数 " } ], [ { "aoVal": "C", "content": "$$\\left( \\frac{1}{{{a}_{1}}}+1 \\right)\\left( \\frac{1}{{{a}_{2}}}+2 \\right)\\cdots \\left( \\frac{1}{{{a}_{n}}}+n \\right)$$是正数 " } ], [ { "aoVal": "D", "content": "$$\\left( 1-\\frac{1}{{{a}_{1}}} \\right)\\left( 2-\\frac{1}{{{a}_{2}}} \\right)\\cdots \\left( n-\\frac{1}{{{a}_{n}}} \\right)$$是正数 " } ] ]

[ "课内体系->能力->运算能力", "课内体系->能力->抽象概括能力", "课内体系->知识点->数->有理数->有理数基础运算->有理数乘方->含乘方的有理数混合运算", "课内体系->知识点->综合与实践->规律探究与程序框图->规律探究->算式找规律" ]

[ "$${{a}_{1}}$$，$${{a}_{2}}$$，\\ldots，$${{a}_{n}}$$是$$n$$个互不相同的负整数，其中$$n$$是奇自然数． 若$${{a}_{1}}=-1$$，$$({{a}_{1}}+1)=0$$，则$$({{a}_{1}}+1)({{a}_{2}}+2)\\cdots ({{a}_{n}}+n)=0$$，排除$$\\text{A}$$． 若$${{a}_{1}}=-1$$，$${{a}_{2}}=-2$$，$${{a}_{3}}=-3$$，\\ldots，$${{a}_{n}}=-n$$时， $$({{a}_{1}}-1)({{a}_{2}}-2)\\cdots ({{a}_{n}}-n)=(-2)(-4)(-6)\\cdots (-2n)$$$$={{(-1)}^{n}}2\\times 4\\times 6\\times \\cdots \\times (2n) ~\\textless{} ~0$$（因为$$n$$是奇数）．故排除$$\\text{B}$$． 若$${{a}_{1}}=-1$$时，$$\\left( \\frac{1}{{{a}_{1}}}+1 \\right)=0$$， 故$$\\left( \\frac{1}{{{a}_{1}}}+1 \\right)\\left( \\frac{1}{{{a}_{2}}}+2 \\right)\\cdots \\left( \\frac{1}{{{a}_{n}}}+n \\right)=0$$， 排除$$\\text{C}$$．所以应选$$\\text{D}$$． 事实上，若$${{a}_{1}} ~\\textless{} ~0$$，$${{a}_{2}} ~\\textless{} ~0$$，\\ldots，$${{a}_{n}} ~\\textless{} ~0$$，则$$-\\frac{1}{{{a}_{1}}}\\textgreater0$$，$$-\\frac{1}{{{a}_{2}}}\\textgreater0$$，\\ldots，$$-\\frac{1}{{{a}_{n}}}\\textgreater0$$． 所以$$1-\\frac{1}{{{a}_{1}}}\\textgreater0$$，$$2-\\frac{1}{{{a}_{2}}}\\textgreater0$$，\\ldots，$$n-\\frac{1}{{{a}_{n}}}\\textgreater0$$． 所以$$\\left( 1-\\frac{1}{{{a}_{1}}} \\right)\\left( 2-\\frac{1}{{{a}_{2}}} \\right)\\cdots \\left( n-\\frac{1}{{{a}_{n}}} \\right)\\textgreater0$$，选$$\\text{D}$$． " ]

[ "2017~2018学年福建泉州丰泽区泉州实验中学（圣湖校区）初一下学期期中第8题4分", "1995年第6届全国希望杯初一竞赛复赛第10题", "2015~2016学年北京海淀区清华大学附属中学初一下学期期中第10题3分" ]

[ [ { "aoVal": "A", "content": "$$80$$分 " } ], [ { "aoVal": "B", "content": "$$76$$分 " } ], [ { "aoVal": "C", "content": "$$75$$分 " } ], [ { "aoVal": "D", "content": "$$64$$分 " } ] ]

[ "课内体系->知识点->方程与不等式->不等式（组）->不等式组应用->不等式组的其他实际问题" ]

[ "设在模拟考试中至少要得$$x$$分，则在模拟考试中至少做对$$\\frac{x}{4}$$道题， 做错或不会做的题至多是$$25-\\frac{x}{4}$$道题． 在正式考试中要出现模拟考试中$$80$$分的试题，即$$\\frac{80}{4}=20$$道题． 如果最坏的可能，即其余$$20$$分题（$$5$$道新题）某人全不会做， 而且模拟考试中$$25-\\frac{x}{4}$$道失分的题又全出现在正式考试试题之中， 并且该生在模拟考试后也没能复习纠错，仍按错误答案在正规考试中失分， 这时该生只能从$$\\frac{80}{4}-(25-\\frac{x}{4})$$道题中取得及格分， 即$$\\left[ \\frac{80}{4}-(25-\\frac{x}{4}) \\right]\\times 4\\geqslant 60$$，解得$$x\\geqslant 80$$． 即某人欲在正式考试中确保及格，则他在模拟考试中至少要得$$80$$分．选$$\\text{A}$$． " ]

[ "2011年第22届全国希望杯初二竞赛初赛第4题4分" ]

[ [ { "aoVal": "A", "content": "$$2$$个 " } ], [ { "aoVal": "B", "content": "$$3$$个 " } ], [ { "aoVal": "C", "content": "$$4$$个 " } ], [ { "aoVal": "D", "content": "$$5$$个 " } ] ]

[ "课内体系->知识点->函数->一次函数->一次函数基础" ]

[ "解方程组$$\\left { \\begin{matrix}y=2x-1 y=kx+k \\end{matrix} \\right.$$， 得$$x=\\frac{-1-k}{k-2}=-1-\\frac{3}{k-2}$$． 因为交点的横坐标$$x$$的值为整数（此时交点的纵坐标$$y$$也一定为整数）， 则$$k-2$$应该是$$3$$的约数， 即$$k-2$$的值可以是$$1$$，$$-1$$，$$3$$，$$-3$$， 所以$$k$$的值可以是$$3$$，$$1$$，$$5$$，$$-1$$，共有$$4$$个． " ]

[ "2001年第12届希望杯初一竞赛第10题", "初一上学期单元测试《一元一次方程》第20题" ]

[ [ { "aoVal": "A", "content": "$$4$$个 " } ], [ { "aoVal": "B", "content": "$$8$$个 " } ], [ { "aoVal": "C", "content": "$$12$$个 " } ], [ { "aoVal": "D", "content": "$$16$$个 " } ] ]

[ "课内体系->知识点->方程与不等式->一元一次方程->含参一元一次方程->一元一次方程的含参整数解" ]

[ "$$x=\\frac{2001}{k+1}$$为整数，又因为$$2001=1\\times 3\\times 23\\times 29$$，$$k+1$$可取$$\\pm 1$$、$$\\pm 3$$、$$\\pm 23$$、$$\\pm 29$$、$$\\pm (3\\times 23)$$、$$\\pm (3\\times 29)$$、$$\\pm (23\\times 29)$$、$$\\pm 2001$$共$$16$$个值，对应的$$k$$值也有$$16$$个． " ]

[ "2012年第23届全国希望杯初一竞赛初赛第9题4分" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$36$$ " } ] ]

[ "课内体系->知识点->三角形->三角形及多边形->与三角形有关的线段->判断三角形的高，中线和角平分线", "课内体系->知识点->几何图形初步->直线、射线、线段->直线、射线、线段的运算->线段中点与等分点" ]

[ "因为$$AD:DC=1:3$$，且$${{S}_{\\triangle ABC}}=60$$， 所以$${{S}_{\\triangle BCD}}=\\frac{3}{4}{{S}_{\\triangle ABC}}=45$$． 因为$$BE:ED=4:1$$， 所以$${{S}_{\\triangle BCE}}=\\frac{4}{5}{{S}_{\\triangle BCD}}=\\frac{4}{5}\\times 45=36$$． 因为$$EF:FC=4:5$$， 所以$${{S}_{\\triangle BEF}}=\\frac{4}{9}{{S}_{\\triangle BCE}}=\\frac{4}{9}\\times 36=16$$． " ]

[ "2010年第21届全国希望杯初一竞赛初赛第5题4分" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$35$$ " } ], [ { "aoVal": "C", "content": "$$40$$ " } ], [ { "aoVal": "D", "content": "$$60$$ " } ] ]

[ "课内体系->知识点->数->有理数->有理数与实际问题->有理数乘除法与实际问题" ]

[ "总工程量的$$\\frac{1}{3}$$为$$8\\times 35=280$$（人$$\\cdot $$天）， 还有$$\\frac{2}{3}$$，即$$2\\times 280=560$$（人$$\\cdot $$天）， 要完成剩余的工程，还需要$$\\frac{560}{8+6}=40$$（天）． " ]

[ "2009年第20届希望杯初一竞赛第1试第3题4分" ]

[ [ { "aoVal": "A", "content": "$$2006$$个 " } ], [ { "aoVal": "B", "content": "$$2007$$个 " } ], [ { "aoVal": "C", "content": "$$2008$$个 " } ], [ { "aoVal": "D", "content": "$$2009$$个 " } ] ]

[ "课内体系->知识点->数->有理数->数轴->数轴上整点覆盖问题", "课内体系->能力->运算能力" ]

[ "当画的线段的两个端点都在整点上时，线段$$AB$$盖住$$2009$$个整点；当画的线段的两个端点都不在整点上时，线段$$AB$$盖住$$2008$$个整点；所以长为$$2008$$厘米的线段$$AB$$，至少可盖住数轴上$$2008$$个整点． 故选$$\\text{C}$$． " ]

[ "2015年第32届全国全国初中数学联赛竞赛A卷第6题7分" ]

[ [ { "aoVal": "A", "content": "$$285$$ " } ], [ { "aoVal": "B", "content": "$$350$$ " } ], [ { "aoVal": "C", "content": "$$540$$ " } ], [ { "aoVal": "D", "content": "$$635$$ " } ] ]

[ "课内体系->知识点->数->有理数->有理数基础运算->有理数除法->有理数四则混合运算", "课内体系->能力->运算能力", "竞赛->知识点->数论->同余->剩余系及其应用", "竞赛->知识点->数论->整除->整除的概念与基本性质" ]

[ "∵$$5{{n}^{2}}+3n-5$$是$$15$$的倍数， ∴$$5\\textbar(5{{n}^{2}}+3n-5)$$， ∴$$5\\textbar3n$$， ∴$$5\\textbar n$$． 设$$n=5m$$（$$m$$是正整数）， 则$$5{{n}^{2}}+3n-5=125{{m}^{2}}+15m-5=120{{m}^{2}}+15m+5({{m}^{2}}-1)$$． 又∵$$5{{n}^{2}}+3n-5$$是$$15$$的倍数， ∴$${{m}^{2}}-1$$是$$3$$的倍数， ∴$$m=3k+1$$或$$m=3k+2$$，其中$$k$$是非负整数， ∴$$n=5(3k+1)=15k+5$$或$$n=5(3k+2)=15k+10$$，其中$$k$$是非负整数， ∴符合条件的所有正整数$$n$$的和为 $$(5+20+35+50+65+80+95)+(10+25+40+55+70+85)=635$$． " ]

[ "2016年第27届全国希望杯初一竞赛初赛第2题4分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{mn}{m+n}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{m-n}{m+n}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{m+n}{mn}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{mn}{m+2n}$$ " } ] ]

[ "课内体系->知识点->式->整式的加减->整式有关的概念->代数式->列代数式", "课内体系->能力->分析和解决问题能力" ]

[ "$$n$$个人完成一项工程需要$$m$$天， 则$$1$$个人完成这项工程需要$$mn$$天， ∴$$(m+n)$$个人完成这项工程需要$$\\frac{mn}{m+n}$$天． " ]

[ "1995年第6届全国希望杯初一竞赛初赛第3题" ]

[ [ { "aoVal": "A", "content": "$${{a}^{2}}={{(-a)}^{2}}$$ " } ], [ { "aoVal": "B", "content": "$${{a}^{3}}={{(-a)}^{3}}$$ " } ], [ { "aoVal": "C", "content": "$${{a}^{2}}=\\left\\textbar{} {{a}^{2}} \\right\\textbar$$ " } ], [ { "aoVal": "D", "content": "$${{a}^{3}}=-\\left\\textbar{} {{a}^{3}} \\right\\textbar$$ " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->数->有理数->绝对值->绝对值的非负性", "课内体系->知识点->数->有理数->绝对值->绝对值的定义", "课内体系->知识点->数->有理数->有理数基础运算->有理数乘法->有理数乘法运算", "课内体系->知识点->数->有理数->有理数基础运算->有理数乘方->有理数乘方运算" ]

[ "以特殊值$$a=-2$$代入检验，易知$${{(-2)}^{2}}={{(-(-2))}^{2}}$$，$${{(-2)}^{3}}\\ne {{(-(-2))}^{3}}$$，$${{(-2)}^{2}}=\\left\\textbar{} {{(-2)}^{2}} \\right\\textbar$$，$${{(-2)}^{3}}=-\\left\\textbar{} {{(-2)}^{3}} \\right\\textbar$$，所以选$$\\text{B}$$． " ]

[ "1993年第10届全国初中数学联赛竞赛第1题" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$-1$$ " } ], [ { "aoVal": "C", "content": "$$x-1$$ " } ], [ { "aoVal": "D", "content": "$$x+1$$ " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->式->整式的乘除->整式的乘除运算->综合除法和余数定理 ", "课内体系->知识点->式->整式的乘除->整式的乘除运算->大除法" ]

[ "直接利用整式除法比较繁琐，将原式变形得 $${{x}^{12}}-{{x}^{6}}+1={{x}^{6}}({{x}^{6}}-1)+1$$， $$={{x}^{6}}({{x}^{2}}-1)({{x}^{4}}+{{x}^{2}}+1)+1$$． 所以余式为$$1$$． " ]

[ "2014年第25届全国希望杯初二竞赛初赛第4题4分", "2016~2017学年安徽安庆潜山市潜山第四中学初三上学期期末模拟第2题4分", "2018~2019学年广东深圳福田区深圳实验学校中学部初三上学期开学考试第4题3分", "2015~2016学年江苏苏州相城区初二下学期期中第7题3分", "2018年陕西西安碑林区西安市第六中学初三中考一模第6题3分", "2016~2017学年12月北京海淀区北方交通大学附属中学分校初三上学期月考第6题3分", "2017~2018学年10月山东济南历下区济南燕山学校初三上学期月考第10题3分", "2016~2017学年安徽合肥肥西县肥西县烧脉中学初三上学期期中第4题4分" ]

[ [ { "aoVal": "A", "content": "$${{y}_{1}}\\textless{}{{y}_{2}}\\textless{}{{y}_{3}}$$ " } ], [ { "aoVal": "B", "content": "$${{y}_{2}}\\textless{}{{y}_{1}}\\textless{}{{y}_{3}}$$ " } ], [ { "aoVal": "C", "content": "$${{y}_{3}}\\textless{}{{y}_{1}}\\textless{}{{y}_{2}}$$ " } ], [ { "aoVal": "D", "content": "$${{y}_{3}}\\textless{}{{y}_{2}}\\textless{}{{y}_{1}}$$ " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->函数->反比例函数->反比例函数图象与性质->反比例函数增减性" ]

[ "∵反比例函数$$y=\\frac{6}{x}$$中，$$k=6\\textgreater0$$， ∴此反比例函数图象的两个分支在一、三象限， ∵$${{x}_{3}}\\textgreater0$$， ∴点$$\\left( {{x}_{3}},{{y}_{3}} \\right)$$在第一象限，$${{y}_{3}}\\textgreater0$$， ∵$${{x}_{1}} ~\\textless{} ~{{x}_{2}} ~\\textless{} ~0$$， ∴点$$\\left( {{x}_{1}},{{y}_{1}} \\right)$$，$$\\left( {{x}_{2}},{{y}_{2}} \\right)$$在第三象限，$$y$$随$$x$$的增大而减小，故$${}{{y}_{2}} ~\\textless{} {} {{y}_{1}}\\textless0$$，于是$${{y}_{2}} ~\\textless{} ~{{y}_{1}} ~\\textless{} ~{{y}_{3}}$$． 故选$$\\text{B}$$． " ]

[ "2018年重庆中考真题A卷第12题4分", "2018年四川绵阳涪城区绵阳东辰国际学校初三自主招生第3题4分", "2019~2020学年重庆合川区西南大学银翔实验中学初一下学期期末模拟（一）第12题4分", "2018~2019学年广东深圳南山区桃苑学校初二下学期期末第11题3分", "2019年浙江杭州拱墅区杭州市文澜中学初二竞赛第7题3分", "2019~2020学年江苏苏州姑苏区苏州市振华中学校初三下学期单元测试《方程与不等式》第10题3分" ]

[ [ { "aoVal": "A", "content": "$$-3$$ " } ], [ { "aoVal": "B", "content": "$$-2$$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "$$2$$ " } ] ]

[ "课内体系->知识点->方程与不等式->分式方程->分式方程的解与解分式方程->分式方程与不等式综合", "课内体系->能力->运算能力" ]

[ "$$\\begin{cases}\\dfrac{x-1}{2}\\textless{}\\dfrac{1+x}{3} 5x-2\\geqslant x+a \\end{cases}$$， 不等式组整理得：$$\\begin{cases}x\\textless{}5 x\\geqslant \\dfrac{a+2}{4} \\end{cases}$$， 由不等式组有且只有四个整数解，得到$$0\\textless{}\\frac{a+2}{4}\\leqslant 1$$， 解得：$$-2\\textless{}a\\leqslant 2$$，即整数$$a=-1$$，$$0$$，$$1$$，$$2$$， $$\\frac{y+a}{y-1}+\\frac{2a}{1-y}=2$$， 分式方程去分母得：$$y+a-2a=2(y-1)$$， 解得：$$y=2-a$$， 由分式方程的解为非负数以及分式有意义的条件，得到$$a$$为$$-1$$，$$0$$，$$2$$，之和为$$1$$． " ]

[ "2017~2018学年浙江宁波鄞州区宁波市鄞州蓝青学校初二下学期期末第7题3分", "1991年第8届全国初中数学联赛竞赛（第一试）第1题" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "~$$\\frac{1}{3}$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{5}{3}$$ " } ] ]

[ "课内体系->知识点->式->分式->分式化简求值->分式化简求值-条件化简求值", "课内体系->知识点->数->实数->平方根->算术平方根的双重非负性", "课内体系->能力->运算能力" ]

[ "据算术根性质，由右端知$$y\\leqslant a\\leqslant x$$， 又由左端知$$a\\geqslant 0$$且$$a\\leqslant 0$$，故$$a=0$$． 由此得$$x=-y$$，代入所求式算值为$$\\frac{1}{3}$$． 故选$$\\text{B}$$． " ]

[ "1996年第7届全国希望杯初一竞赛复赛第1题" ]

[ [ { "aoVal": "A", "content": "$$-{{\\left( -a \\right)}^{2}}$$ " } ], [ { "aoVal": "B", "content": "$$-\\left\\textbar{} -a \\right\\textbar$$ " } ], [ { "aoVal": "C", "content": "$$-{{a}^{2}}$$ " } ], [ { "aoVal": "D", "content": "$$-\\left( -{{a}^{2}} \\right)$$ " } ] ]

[ "课内体系->知识点->数->有理数->正数和负数->正数、负数定义" ]

[ "当$$a\\textless{}0$$时，$${{\\left( -a \\right)}^{2}}\\textgreater0$$，$$\\left\\textbar{} -a \\right\\textbar\\textgreater0$$，$${{a}^{2}}\\textgreater0$$， 所以$$-{{\\left( -a \\right)}^{2}}\\textless{}0$$，$$-\\left\\textbar{} -a \\right\\textbar\\textless{}0$$，$$-{{a}^{2}}\\textless{}0$$，因此排除$$\\text{A}$$、$$\\text{B}$$、$$\\text{C}$$，选$$\\text{D}$$． 事实上，$$a\\textless{}0$$时，$${{a}^{2}}\\textgreater0$$，$$-\\left( -{{a}^{2}} \\right)\\textgreater0$$．当然$$a=-0.01$$时更是如此． " ]

[ "2017~2018学年浙江金华金东区初二上学期期末第10题3分", "2017~2018学年9月安徽安庆潜山市潜山第四中学初二上学期月考第6题4分", "2017~2018学年浙江宁波镇海区宁波市镇海蛟川书院初二上学期期中第9题4分", "2017~2018学年12月陕西西安碑林区西北工业大学附属中学初二上学期月考第7题3分", "2018~2019学年辽宁大连甘井子区初一下学期期末第10题3分", "2018年湖南长沙初二竞赛长郡教育集团（觉园杯）第1题4分", "2018~2019学年浙江宁波镇海区宁波市镇海蛟川书院初二上学期期中第4题4分", "2017年广西贵港中考真题第6题3分", "2020~2021学年4月浙江杭州上城区杭州市建兰中学初三下学期周测A卷第1题4分", "2018~2019学年湖北武汉武昌区初一下学期期中（十五校联考）第8题3分" ]

[ [ { "aoVal": "A", "content": "第一象限 " } ], [ { "aoVal": "B", "content": "第二象限 " } ], [ { "aoVal": "C", "content": "第三象限 " } ], [ { "aoVal": "D", "content": "第四象限 " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->函数->平面直角坐标系->坐标系基础->象限内坐标的特征", "课内体系->思想->分类讨论思想" ]

[ "①$$m-3\\textgreater0$$，即$$m\\textgreater3$$时， $$-2m\\textless{}-6$$，$$4-2m\\textless{}-2$$， 所以，点$$P(m-3,4-2m)$$在第四象限，不可能在第一象限． ②$$m-3\\textless{}0$$，即$$m\\textless{}3$$时， $$-2m\\textgreater-6$$，$$4-2m\\textgreater-2$$， 点$$P(m-3,4-2m)$$可能在第二或三象限， 综上所述，点$$P$$不可能在第一象限． " ]

[ "1991年第8届全国初中数学联赛竞赛（第一试）第4题" ]

[ [ { "aoVal": "A", "content": "$${{1991}^{-1}}$$ " } ], [ { "aoVal": "B", "content": "$$-{{1991}^{-1}}$$ " } ], [ { "aoVal": "C", "content": "$${{\\left( -1 \\right)}^{n}}1991$$ " } ], [ { "aoVal": "D", "content": "$${{\\left( -1 \\right)}^{n}}{{1991}^{-1}}$$ " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->式->二次根式->二次根式化简求值", "课内体系->知识点->式->整式的乘除->乘法公式->完全平方公式的计算-不含分式" ]

[ "$$1+{{x}^{2}}=1+\\frac{1}{4}\\left( {{1991}^{\\frac{2}{n}}}-2+{{1991}^{-\\frac{2}{n}}} \\right)=\\frac{1}{4}{{\\left( {{1991}^{\\frac{1}{n}}}+{{1991}^{-\\frac{1}{n}}} \\right)}^{2}}$$， 原式$$={{\\left[ \\frac{1}{2}\\left( {{1991}^{\\frac{1}{n}}}-{{1991}^{-\\frac{1}{n}}} \\right)-\\frac{1}{2}\\left( {{1991}^{\\frac{1}{n}}}+{{1991}^{-\\frac{1}{n}}} \\right) \\right]}^{n}}={{\\left( -1 \\right)}^{n}}{{1991}^{-1}}$$． " ]

[ "1999年第10届希望杯初二竞赛第2试第1题" ]

[ [ { "aoVal": "A", "content": "①，②，③ " } ], [ { "aoVal": "B", "content": "②，③，④ " } ], [ { "aoVal": "C", "content": "③，④，⑤ " } ], [ { "aoVal": "D", "content": "①，②，④ " } ] ]

[ "课内体系->知识点->式->因式分解->公式法->提公因式+平方差", "课内体系->知识点->式->因式分解->提公因式法" ]

[ "②式$$={{(x-3a)}^{3}}$$， ③式$$=x(b+c-d)+y(b+c-d)-2(b+c-d)$$ $$=(b+c-d)(x+y-2)$$， ④式$$=(m-n)(3m-6n)=3(m-n)(m-2n)$$． 所以，②、③、④式合乎要求． 故选$$\\text{B}$$． " ]

[ "2008年第19届希望杯初二竞赛第2试第6题" ]

[ [ { "aoVal": "A", "content": "$$2$$或$$4$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$4$$或$$8$$ " } ], [ { "aoVal": "D", "content": "$$2$$到$$46$$之间的任意偶数 " } ] ]

[ "竞赛->知识点->方程与不等式->一次方程->方程组" ]

[ "设$$10$$元、$$20$$元、$$50$$元分别有$$x$$，$$y$$，$$24-\\left( x+y \\right)$$张， 则$$10x+20y+50\\left( 24-x-y \\right)=1000$$， 即$$40x+30y=200$$，$$4x+3y=20$$， 其中$$x$$，$$y$$都是正整数，由$$x\\geqslant 1$$知$$3y\\leqslant 16$$， 所以$$1\\leqslant y\\leqslant \\frac{16}{3}$$， 所以，$$y$$只能从$$1$$，$$2$$，$$3$$，$$4$$，$$5$$中取， 又$$3y=4\\left( 5-x \\right)$$，其中$$5-x$$是正整数，$$3$$与$$4$$是互质的， 所以，$$y$$中一定有一个因数$$4$$， 因此$$y$$只能取$$4$$， 故选$$\\text{B}$$． " ]