Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "2012年第23届全国希望杯初一竞赛复赛第6题4分" ]

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

[ "竞赛->知识点->数与式->数的运算->有理数运算问题" ]

[ "$$77=\\frac{11\\times 7\\times 2}{2}$$． $$77=\\frac{(11\\times 7)\\times 2}{2}=\\frac{(38+39)\\times 2}{2}=38+39$$； $$77=\\frac{(11\\times 2)\\times 7}{2}=\\frac{(8+14)\\times 7}{2}=8+9+10+11+12+13+14$$； $$77=\\frac{(2\\times 7)\\times 11}{2}=\\frac{(2+12)\\times 11}{2}=2+3+4+5+6+7+8+9+10+11+12$$． 综上，$$n$$的值的个数是$$3$$． " ]

[ "2020~2021学年6月江苏苏州工业园区星港学校初一下学期周测B卷第8题2分", "2003年第14届希望杯初二竞赛第2试第1题", "2020~2021学年江苏苏州高新区苏州市高新区实验初级中学初二下学期期末模拟（一）第7题" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$-1$$ " } ], [ { "aoVal": "C", "content": "$$1$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->式->因式分解->其他方法->待定系数法" ]

[ "由因式定理可知，当$$y-2x+1=0$$时， $$4xy-4{{x}^{2}}-{{y}^{2}}-k=0$$， 因此可令$$\\begin{cases}x=0 y=-1 \\end{cases}$$， 则$$4xy-4{{x}^{2}}-{{y}^{2}}-k=-1-k=0$$， ∴$$k=-1$$． 所以答案为：$$k=-1$$． " ]

[ "2017~2018学年江苏无锡滨湖区无锡外国语学校初一下学期期中第10题3分", "2007年竞赛第1题6分" ]

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

[ "知识标签->学习能力->运算能力", "知识标签->数学思想->分类讨论思想", "知识标签->题型->方程与不等式->二元一次方程（组）->解二元一次方程组->题型：解含绝对值的方程组", "知识标签->知识点->方程与不等式->二元一次方程（组）->二元一次方程组->二元一次方程组的解", "知识标签->知识点->方程与不等式->二元一次方程（组）->二元一次方程组->加减消元法解二元一次方程组" ]

[ "①当$$x\\textgreater0$$，$$y\\textgreater0$$时，原方程组为$$\\begin{cases}x+y=12 x+y=6 \\end{cases}$$显然无解； ②当$$x ~\\textless{} ~0$$，$$y\\textgreater0$$时，原方程组为$$\\begin{cases}-x+y=12 x+y=6 \\end{cases}$$，解得$$\\begin{cases}x=-3 y=9 \\end{cases}$$； ③当$$x\\textgreater0$$，$$y ~\\textless{} ~0$$时，原方程组为$$\\begin{cases}x+y=12 x-y=6 \\end{cases}$$，解得$$\\begin{cases}x=9 y=3 \\end{cases}$$，舍去； ④当$$x ~\\textless{} ~0$$，$$y ~\\textless{} ~0$$时，原方程组为$$\\begin{cases}-x+y=12 x-y=6 \\end{cases}$$显然无解； 综上，只有$$1$$组解．故选$$\\text{A}$$． 若$$x\\geqslant 0$$，则$$\\begin{cases}x+y=12 x+\\left\\textbar{} y \\right\\textbar=6 \\end{cases}$$，于是$$\\left\\textbar{} y \\right\\textbar-y=-6$$，显然不可能． 若$$x ~\\textless{} ~0$$，则$$\\begin{cases}-x+y=12 x+\\left\\textbar{} y \\right\\textbar=6 \\end{cases}$$， 于是$$\\left\\textbar{} y \\right\\textbar+y=18$$，解得$$y=9$$，进而求得$$x=-3$$． 所以，原方程组的解为$$\\begin{cases}x=-3 y=9 \\end{cases}$$，只有$$1$$个解． 故选$$\\text{A}$$． " ]

[ "1990年第1届全国希望杯初一竞赛初赛第7题" ]

[ [ { "aoVal": "A", "content": "$$a$$大于$$-a$$ " } ], [ { "aoVal": "B", "content": "$$a$$小于$$-a$$ " } ], [ { "aoVal": "C", "content": "$$a$$大于$$-a$$或$$a$$小于$$-a$$ " } ], [ { "aoVal": "D", "content": "$$a$$不一定大于$$-a$$ " } ] ]

[ "课内体系->知识点->数->有理数->相反数->相反数的性质" ]

[ "令$$a=0$$，马上可以排除$$\\text{A}$$、$$\\text{B}$$、$$\\text{C}$$，应选$$\\text{D}$$． " ]

[ "2019年全美数学竞赛（AMC）竞赛第15题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{14}{85}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{7}{25}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{2}{5}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{4}{7}$$ " } ], [ { "aoVal": "E", "content": "$$\\frac{7}{10}$$ " } ] ]

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

[ "海滩上有$$50$$人戴墨镜，有$$35$$人戴帽子．而有些人既戴墨镜又戴帽子．如果从戴帽子的人里随机选一个，选中戴墨镜的概率是$$\\frac{2}{5}$$．相反地，如果从戴墨镜的人里随机选一个，这个人戴帽子的概率是多大？ This is a problem of conditional probability, a good application of Bayes formula. $$P(G\\textbar C)=\\frac{2}{5}$$， $$\\textbar G\\cap C\\textbar=\\textbar C\\textbar\\cdot P(G\\textbar C)=14$$， $$P(C\\textbar G)=\\frac{\\textbar G\\cap C\\textbar}{\\textbar G\\textbar}=\\frac{7}{25}$$． " ]

[ "1993年第4届希望杯初二竞赛第5题" ]

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

[ "竞赛->知识点->数与式->因式分解->因式分解：公式法" ]

[ "等式$$2x+{{x}^{2}}+{{x}^{2}}{{y}^{2}}+2=-2xy$$即$${{(x+1)}^{2}}+{{(xy+1)}^{2}}=0$$， 所以$$x+1=0$$，$$xy+1=0$$， 解得$$x=-1$$，$$y=1$$，则$$x+y=0$$． 故选$$\\text{B}$$． " ]

[ "2010年第21届全国希望杯初一竞赛初赛第7题4分" ]

[ [ { "aoVal": "A", "content": "$$\\left( a-1 \\right)\\left( a-2 \\right)\\textless{}0$$ " } ], [ { "aoVal": "B", "content": "$$\\left( a-1 \\right)\\left( a-2 \\right)\\textgreater0$$ " } ], [ { "aoVal": "C", "content": "$$\\left( a-3 \\right)\\left( a-4 \\right)\\textless{}0$$ " } ], [ { "aoVal": "D", "content": "$$\\left( a-3 \\right)\\left( a-4 \\right)\\textgreater0$$ " } ] ]

[ "课内体系->知识点->方程与不等式->一元一次方程->一元一次方程的相关概念->一元一次方程的解", "课内体系->知识点->方程与不等式->一元一次方程->一元一次方程的相关概念->由一元一次方程的解求参数的值" ]

[ "由$$x-2a+4=0$$，得$$x=2a-4=2(a-2)$$， 因为方程$$x-2a+4=0$$的根是负数， 所以$$a\\textless{}2$$， 所以$$a-3\\textless{}0$$，$$a-4\\textless{}0$$， 所以$$\\left( a-3 \\right)\\left( a-4 \\right)\\textgreater0$$． " ]

[ "1999年竞赛（全国初中数学竞赛）第6题5分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$19$$ " } ], [ { "aoVal": "D", "content": "$$26$$ " } ] ]

[ "课内体系->知识点->方程与不等式->其他方程->多元二次方程（组）", "课内体系->能力->运算能力" ]

[ "由$${{x}^{3}}+{{y}^{3}}+3xyz={{z}^{3}}$$， 得$${{x}^{3}}+{{y}^{3}}+{{\\left( -z \\right)}^{3}}-3xy\\left( -z \\right)=0$$， 所以$$\\left( x+y-z \\right)\\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy+yz+zx \\right)=0$$， 因为，$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}-xy+yz+zx=\\frac{1}{2}\\left[ {{\\left( x-y \\right)}^{2}}+{{\\left( y+z \\right)}^{2}}+{{\\left( z+x \\right)}^{2}} \\right]\\textgreater0$$， 所以$$x+y-z=0$$，即$$z=x+y$$， 所以$$7{{x}^{2}}={{z}^{2}}-{{x}^{2}}=\\left( z-x \\right)\\left( z+x \\right)=y\\left( 2z+y \\right)$$， ∴$$7y=2x+y$$， 所以$$x=3y$$，$$z=4y$$， 由于$$1=\\left( x,y \\right)=\\left( 3y,y \\right)=y$$， 所以，$$x=3$$，$$y=1$$，$$z=4$$，于是$${{x}^{2}}+{{y}^{2}}+{{z}^{2}}=26$$． 故选：$$\\text{D}$$． " ]

[ "2012年第29届全国全国初中数学联赛竞赛第7题7分" ]

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

[ "知识标签->题型->方程与不等式->其它方程->一次不定方程", "知识标签->知识点->方程与不等式->不定方程" ]

[ "由$$a+\\frac{1}{b}=b+\\frac{1}{c}=c+\\frac{1}{a}=t$$， 得$$b=\\frac{1}{t-a}$$，$$c=\\frac{at-1}{a}$$，$$\\frac{1}{t-a}+\\frac{at-1}{a}=t$$， 整理得$$a{{t}^{3}}-{{t}^{2}}-{{a}^{2}}{{t}^{2}}+{{a}^{2}}-at+1=0$$， 即$$({{t}^{2}}-1)({{a}^{2}}-at+1)=0$$． 同理得：$$({{t}^{2}}-1)({{b}^{2}}-bt+1)=0$$，$$({{t}^{2}}-1)({{c}^{2}}-ct+1)=0$$． 若$${{t}^{2}}-1\\ne 0$$，则$$a$$，$$b$$，$$c$$为二次方程$${{x}^{2}}-xt+1=0$$的解， 这与$$a$$，$$b$$，$$c$$互不相等矛盾，不满足题意， 故$${{t}^{2}}-1=0$$， 即$$t=\\pm 1$$． $$\\frac{1}{b}=t-a$$，∴$$b=\\frac{1}{t-a}$$， $$c=t-\\frac{1}{a}$$，∴$$\\frac{1}{c}=\\frac{1}{t-\\dfrac{1}{a}}$$， ∴$$t=b+\\frac{1}{c}=\\frac{1}{t-a}+\\frac{1}{t-\\dfrac{1}{a}}$$， ∴$$t\\left( t-a \\right)\\left( t-\\dfrac{1}{a} \\right)=t-\\dfrac{1}{a}+t-a$$， 整理得$$\\left( t-a-\\frac{1}{a} \\right)\\left( {{t}^{2}}-1 \\right)=0$$， $$a$$，$$ b$$，$$ c$$互不相等，∴$$t\\ne a+\\frac{1}{a}$$， ∴$${{t}^{2}}=1$$， ∴$$t=\\pm 1$$． 故答案为：$$\\pm 1$$． " ]

[ "2018年全国初中数学联赛初一竞赛" ]

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

[ "课内体系->知识点->式->整式的加减->整式有关的概念->多项式", "竞赛->知识点->数与式->整式->对称多项式 " ]

[ "不要搜题 " ]

[ "1997年第8届希望杯初二竞赛第1试第4题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{4}$$或$$\\frac{3}{4}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{3}$$或$$\\frac{2}{3}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{5}$$或$$\\frac{4}{5}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{2}{7}$$或$$\\frac{5}{7}$$ " } ] ]

[ "竞赛->知识点->四边形->特殊平行四边形" ]

[ "两正方形的面积差$$=A{{P}^{2}}-{{\\left( 1-AP \\right)}^{2}}=2AP-1$$， 由题意$$\\left\\textbar{} 2AP-1 \\right\\textbar=\\frac{1}{2}$$，则有 $$\\begin{cases}2AP-1=\\frac{1}{2},\\Rightarrow AP=\\frac{3}{4} 2AP-1=-\\frac{1}{2},\\Rightarrow AP=\\frac{1}{4} \\end{cases}$$． 故选$$\\text{A}$$． " ]

[ "2018年第29届希望杯初一竞赛初赛第1题4分" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2017$$ " } ], [ { "aoVal": "D", "content": "$$2018$$ " } ] ]

[ "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->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->能力->运算能力" ]

[ "$$2017\\times20182018-2018\\times20172017$$ $$=2017\\times2018\\times10001-2018\\times2017\\times10001$$ $$=0$$． " ]

[ "2015年第32届全国全国初中数学联赛竞赛A卷第1题7分" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

[ "知识标签->学习能力->运算能力", "知识标签->题型->式->整式的乘除->乘法公式->题型：利用平方差公式计算", "知识标签->题型->式->分式->分式的运算->题型：分式加减、乘除混合运算", "知识标签->知识点->式->分式->分式的运算->分式的混合运算", "知识标签->知识点->式->整式的乘除->乘法公式->平方差公式" ]

[ "∵$$a+b+c=3$$，$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=4$$， ∴$$\\frac{{{a}^{2}}+{{b}^{2}}}{2-c}+\\frac{{{b}^{2}}+{{c}^{2}}}{2-a}+\\frac{{{c}^{2}}+{{a}^{2}}}{2-b}$$ $$=\\frac{4-{{c}^{2}}}{2-c}+\\frac{4-{{a}^{2}}}{2-a}+\\frac{4-{{b}^{2}}}{2-b}$$ $$=2+c+2+a+2+b$$ $$=a+b+c+6$$ $$=3+6$$ $$=9$$． " ]

[ "2002年第13届希望杯初二竞赛第1试第3题" ]

[ [ { "aoVal": "A", "content": "星期四 " } ], [ { "aoVal": "B", "content": "星期五 " } ], [ { "aoVal": "C", "content": "星期六 " } ], [ { "aoVal": "D", "content": "星期日 " } ] ]

[ "竞赛->知识点->数与式->整式->综合除法和余数定理 " ]

[ "因为$$2001$$年$$7$$月$$13$$日是星期五．又$$365\\div 7=52\\cdots 1$$， $$2004$$年和$$2008$$年均为闰年，比其他几年的$$2$$月份多$$1$$天． 又$$2008$$年距$$2001$$年共$$7$$年时间． 所以，$$2008$$年的$$7$$月$$13$$日应是$$(5+7+2)\\div 7=2\\cdots 0=1\\cdots 7$$， 即$$2008$$年的$$7$$月$$13$$日是星期日． 故选$$\\text{D}$$． " ]

[ "1996年第7届全国希望杯初一竞赛复赛第6题" ]

[ [ { "aoVal": "A", "content": "$$a+b=0$$ " } ], [ { "aoVal": "B", "content": "$$a-b=0$$ " } ], [ { "aoVal": "C", "content": "$$ab=0$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{a}{b}=0$$ " } ] ]

[ "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->解的情况", "课内体系->知识点->方程与不等式->一元一次方程->一元一次方程的相关概念->一元一次方程的解" ]

[ "整理原方程得$$\\left( a+b \\right)x={{a}^{2}}-{{b}^{2}}$$． 要使该方程有无穷多解，只当$$a+b=0$$且$${{a}^{2}}-{{b}^{2}}=0$$， 当$$a+b=0$$时$$a=-b$$，$${{a}^{2}}-{{b}^{2}}=0$$． 所以当$$a+b=0$$时，原方程有无穷多个解，选$$\\text{A}$$． " ]

[ "2019年广东惠州惠城区光正实验学校初二竞赛第1题3分", "2018~2019学年山东日照莒县初二上学期期末第6题3分" ]

[ [ { "aoVal": "A", "content": "$$m\\geqslant -1$$ " } ], [ { "aoVal": "B", "content": "$$m\\textgreater-1$$ " } ], [ { "aoVal": "C", "content": "$$m\\textgreater-1$$且$$m\\ne 3$$ " } ], [ { "aoVal": "D", "content": "$$m\\geqslant -1$$且$$m\\ne 3$$ " } ] ]

[ "课内体系->知识点->式->分式->分式的基础->分式有意义的条件", "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->能力->运算能力" ]

[ "要使式子$$\\frac{\\sqrt{m+1}}{\\textbar m-3\\textbar}$$有意义， ∴$$m+1\\geqslant 0$$且$$m-3\\ne 0$$， ∴$$m\\geqslant -1$$且$$m\\ne 3$$． 选$$\\text{D}$$． " ]

[ "2010年竞赛第4题7分" ]

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

[ "竞赛->知识点->方程与不等式->含参方程" ]

[ "由$${{x}_{1}}=1$$和$${{x}_{k}}={{x}_{k-1}}+1-4\\left( \\left[ \\frac{k-1}{4} \\right]-\\left[ \\frac{k-2}{4} \\right] \\right)$$可得 $${{x}_{1}}=1$$，$${{x}_{2}}=2$$，$${{x}_{3}}=3$$，$${{x}_{4}}=4$$， $${{x}_{5}}=1$$，$${{x}_{6}}=2$$，$${{x}_{7}}=3$$，$${{x}_{8}}=4$$， $$\\cdots $$ 因为$$2010=4\\times 502+2$$，所以$${{x}_{2010}}=2$$． 故选$$\\text{B}$$． " ]

[ "2016年第27届全国希望杯初一竞赛复赛第4题4分" ]

[ [ { "aoVal": "A", "content": "$$36$$ " } ], [ { "aoVal": "B", "content": "$$56$$ " } ], [ { "aoVal": "C", "content": "$$63$$ " } ], [ { "aoVal": "D", "content": "$$65$$ " } ] ]

[ "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->方程与不等式->一元一次方程->一元一次方程与实际问题->一元一次方程的数字问题" ]

[ "设这个数为$$x$$， 由题意，得$$8x-63=\\frac{x}{8}+63$$， 解得$$x=16$$， ∴这道题的正确得数是$$8\\times 16-63=65$$． " ]

[ "2003年第14届希望杯初二竞赛第2试第2题" ]

[ [ { "aoVal": "A", "content": "$$a\\leqslant -\\frac{5}{4}$$ " } ], [ { "aoVal": "B", "content": "$$a\\textless{}-1$$ " } ], [ { "aoVal": "C", "content": "$$-\\frac{5}{4}\\leqslant a ~\\textless{} ~-1$$ " } ], [ { "aoVal": "D", "content": "$$a\\geqslant -\\frac{5}{4}$$ " } ] ]

[ "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" ]

[ "由$$0\\leqslant ax+5\\leqslant 4$$得$$-5\\leqslant ax\\leqslant -1$$． 若$$a\\textgreater0$$，则$$-\\frac{5}{a}\\leqslant x\\leqslant -\\frac{1}{a}$$，$$x$$不可能取到正整数解； 若$$a=0$$，不等式无解． 所以必有$$a\\textless{}0$$，此时$$-\\frac{1}{a}\\leqslant x\\leqslant -\\frac{5}{a}$$． 又因原不等式的整数解是$$1$$、$$2$$、$$3$$、$$4$$， 所以$$0\\textless{}-\\frac{1}{a}\\leqslant 1$$，且$$4\\leqslant -\\frac{5}{a}\\textless{}5$$，解得$$-\\frac{5}{4}\\leqslant a\\textless{}-1$$． " ]

[ "2013年第24届全国希望杯初二竞赛初赛第3题4分" ]

[ [ { "aoVal": "A", "content": "若$$a\\textgreater0$$，则$$a\\textgreater\\frac{1}{a}$$ " } ], [ { "aoVal": "B", "content": "若$$a\\textgreater{{a}^{2}}$$，则$$a\\textgreater1$$． " } ], [ { "aoVal": "C", "content": "若$$0\\textless{}a\\textless{}1$$，则$$a\\textgreater{{a}^{2}}$$ " } ], [ { "aoVal": "D", "content": "若$$\\left\\textbar{} a \\right\\textbar=a$$，则$$a\\textgreater0$$ " } ] ]

[ "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" ]

[ "取特殊值． 当$$a=\\frac{1}{2}$$时，$$a\\textless{}\\frac{1}{a}$$，可知$$\\text{A}$$错误； 当$$a=\\frac{1}{2}\\textless{}1$$时，$$\\frac{1}{2}\\textgreater{{\\left( \\frac{1}{2} \\right)}^{2}}=\\frac{1}{4}$$，可知$$\\text{B}$$错误； 当$$a=0$$时，$$\\left\\textbar{} 0 \\right\\textbar=0$$，可知$$\\text{D}$$错误． 故选$$\\text{C}$$． " ]

[ "2016年第33届全国全国初中数学联赛竞赛第6题7分" ]

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

[ "课内体系->方法->配方法", "课内体系->知识点->式->整式的乘除->整式乘除化简求值->整式乘除化简求值-整体代入化简求值", "课内体系->知识点->式->整式的加减->整式有关的概念->整式->整式的定义", "课内体系->能力->运算能力" ]

[ "方法一：$$M=xy+(2y+3x)z$$ $$=xy+(2y+3x)(1-x-y)$$ $$=-3{{x}^{2}}-4xy-2{{y}^{2}}+3x+2y$$ $$=-2\\left[ {{y}^{2}}+2\\left( x-\\frac{1}{2} \\right)y+{{\\left( x-\\frac{1}{2} \\right)}^{2}} \\right]-3{{x}^{2}}+3x+2{{\\left( x-\\frac{1}{2} \\right)}^{2}}$$ $$=-2{{\\left( y+x-\\frac{1}{2} \\right)}^{2}}-{{x}^{2}}+x+\\frac{1}{2}$$ $$=-2{{\\left( y+x-\\frac{1}{2} \\right)}^{2}}-{{\\left( x-\\frac{1}{2} \\right)}^{2}}+\\frac{3}{4}$$ $$\\leqslant \\frac{3}{4}$$． 当且仅当$$x=\\frac{1}{2}$$，$$y=0$$时，不等式取等号， 所以$${{M}_{\\max }}=\\frac{3}{4}$$． 方法二：令$$y=1-x-z$$，代入$$M$$，则： $$M=x\\left( 1-x-z \\right)+2\\left( 1-x-z \\right)z+3xz=x-{{x}^{2}}-xz+2z-2xz-2{{z}^{2}}+3xz$$ $$=-{{x}^{2}}+x-2{{z}^{2}}+2z=-{{\\left( x-\\frac{1}{2} \\right)}^{2}}-2{{\\left( z-\\frac{1}{2} \\right)}^{2}}+\\frac{3}{4}$$，$${{M}_{\\max }}=\\frac{3}{4}$$． 所以选$$\\text{C}$$． " ]

[ "2016年第27届全国希望杯初一竞赛复赛第1题4分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{2016}$$ " } ], [ { "aoVal": "B", "content": "$$-\\frac{1}{2016}$$ " } ], [ { "aoVal": "C", "content": "$$2016$$ " } ], [ { "aoVal": "D", "content": "$$-2016$$ " } ] ]

[ "课内体系->知识点->数->有理数->相反数->相反数的性质" ]

[ "∵$$3m+5$$与$$m-1$$互为相反数， ∴$$3m+5+m-1=0$$， 解得$$m=-1$$， ∴$$\\frac{\\left\\textbar{} m \\right\\textbar}{2016}=\\frac{1}{2016}$$， ∴$$\\frac{\\left\\textbar{} m \\right\\textbar}{2016}$$的倒数是$$2016$$． " ]

[ "1992年第3届全国希望杯初一竞赛初赛第1题" ]

[ [ { "aoVal": "A", "content": "正整数 " } ], [ { "aoVal": "B", "content": "负整数 " } ], [ { "aoVal": "C", "content": "负分数 " } ], [ { "aoVal": "D", "content": "$$0$$ " } ] ]

[ "课内体系->知识点->数->有理数->有理数的概念及分类->有理数的分类" ]

[ "因$$-\\frac{1}{a}$$为有理数，所以$$a\\ne 0$$． 若$$a=-1$$，$$-\\frac{1}{a}=1$$，排除$$\\text{A}$$； 若$$a=1$$，$$-\\frac{1}{a}=-1$$，排除$$\\text{B}$$； 若$$a=2$$，$$-\\frac{1}{a}=-\\frac{1}{2}$$，排除$$\\text{C}$$．故选$$\\text{D}$$． " ]

[ "1993年第4届全国希望杯初一竞赛初赛第15题" ]

[ [ { "aoVal": "A", "content": "$$-\\frac{1}{328}$$ " } ], [ { "aoVal": "B", "content": "$$-\\frac{1}{329}$$ " } ], [ { "aoVal": "C", "content": "$$-\\frac{1}{337}$$ " } ], [ { "aoVal": "D", "content": "$$-\\frac{1}{340}$$ " } ] ]

[ "课内体系->知识点->数->有理数->倒数与负倒数" ]

[ "前$$15$$个质数为$$2$$，$$3$$，$$5$$，$$7$$，$$11$$，$$13$$，$$17$$，$$19$$，$$23$$，$$29$$，$$31$$，$$37$$，$$41$$，$$43$$，$$47$$． 它们的和为$$328$$，$$328$$的负倒数为$$-\\frac{1}{328}$$，选$$\\text{A}$$． " ]

[ "2009年第20届希望杯初二竞赛第2试第3题" ]

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

[ "课内体系->能力->运算能力", "课内体系->方法->代入法", "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->思想->整体思想" ]

[ "\\textbf{（知识点：多重二次根式）} 因为$$\\sqrt{3\\pm \\sqrt{5}}=\\sqrt{\\frac{6\\pm 2\\sqrt{5}}{2}}=\\frac{\\sqrt{{{\\left( \\sqrt{5}\\pm 1 \\right)}^{2}}}}{\\sqrt{2}}=\\frac{\\sqrt{5}\\pm 1}{\\sqrt{2}}$$， 所以$$x=\\frac{\\sqrt{5}-1}{\\sqrt{2}}-\\frac{\\sqrt{5}+1}{\\sqrt{2}}=\\frac{-2}{\\sqrt{2}}=-\\sqrt{2}\\approx -1.41$$， 所以$$\\left[ x \\right]=-2$$． 所以选$$\\text{A}$$． " ]

[ "1994年第5届全国希望杯初一竞赛复赛第10题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{a}\\textgreater a$$ " } ], [ { "aoVal": "B", "content": "$$\\left\\textbar{} {{a}^{3}} \\right\\textbar\\textgreater{{a}^{3}}$$ " } ], [ { "aoVal": "C", "content": "$$-a\\textgreater{{a}^{2}}$$ " } ], [ { "aoVal": "D", "content": "$${{a}^{3}}\\textless{}-{{a}^{2}}$$ " } ] ]

[ "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" ]

[ "采用特殊值法： 取$$a=-\\frac{1}{2}$$，则$$\\frac{1}{a}=-2$$，$${{a}^{2}}={(-\\frac{1}{2})^{2}}=\\frac{1}{4}$$，$${{a}^{3}}={(-\\frac{1}{2})^{3}}=-\\frac{1}{8}$$． 由$$-2\\textless{}-\\frac{1}{2}$$，排除$$\\text{A}$$； 由$$\\left\\textbar{} -\\frac{1}{8} \\right\\textbar=-(-\\frac{1}{8})$$，排除$$\\text{B}$$； 由$$-\\frac{1}{8}\\textgreater-\\frac{1}{4}$$，排除$$\\text{D}$$． " ]

[ "1992年第3届全国希望杯初一竞赛初赛第10题" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$8\\frac{1}{3}$$ " } ], [ { "aoVal": "C", "content": "$$12\\frac{1}{2}$$ " } ], [ { "aoVal": "D", "content": "$$13$$ " } ] ]

[ "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->知识点->统计与概率->数据的分析->平均数/加权平均数" ]

[ "前三个数之和$$=15\\times 3$$，后两个数之和$$=10\\times 2$$． 所以五个有理数的平均数为$$\\frac{15\\times 3+10\\times 2}{5}=13$$．选$$\\text{D}$$． " ]

[ "2017年第1届重庆全国初中数学联赛初一竞赛初赛" ]

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

[ "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" ]

[ "$$\\because {{\\left( a+b \\right)}^{2}}=6ab$$，$${{\\left( a-b \\right)}^{2}}=2ab$$,且$$a\\textless{}b\\textless{}0$$ $$\\therefore a+b=-\\sqrt{6ab}$$，$$a-b=-\\sqrt{2ab}\\therefore \\frac{a+b}{a-b}=\\sqrt{3}$$ " ]

[ "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", "2018年重庆中考真题A卷第12题4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", "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" ]

[ [ { "aoVal": "A", "content": "$$-3$$ " } ], [ { "aoVal": "B", "content": "$$-2$$ " } ], [ { "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->分式方程->分式方程的解与解分式方程->由分式方程的解确定参数" ]

[ "$$\\begin{cases}\\dfrac{x-1}{2}\\textless{}\\dfrac{1+x}{3} 5x-2\\geqslant x+a \\end{cases}$$， 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$$1$$． " ]

[ "1997年第8届希望杯初二竞赛第2试第10题" ]

[ [ { "aoVal": "A", "content": "$$a\\textgreater b$$ " } ], [ { "aoVal": "B", "content": "$$a\\textless{}b$$ " } ], [ { "aoVal": "C", "content": "$$a=b$$ " } ], [ { "aoVal": "D", "content": "当$$m\\geqslant n$$时，$$a\\geqslant b$$，当$$m\\textless{}n$$时，$$a\\textless{}b$$ " } ] ]

[ "竞赛->知识点->组合->操作与游戏" ]

[ "设训练后回男队的$$10$$人中有$$c$$个女队员，则男队中女队员的人数$$a=c$$．此时，女队中应有男队员的人数为$$b=10-(10-c)=c$$（人）． 所以$$a=b$$． 故选$$\\text{C}$$． " ]

[ "2017~2018学年北京朝阳区朝阳外国语学校初二上学期单元测试《数学基础知识竞赛》第17题3分" ]

[ [ { "aoVal": "A", "content": "$$18{{b}^{2}}-12ab$$ " } ], [ { "aoVal": "B", "content": "$$18{{b}^{2}}+12ab$$ " } ], [ { "aoVal": "C", "content": "$$9{{b}^{2}}-6ab$$ " } ], [ { "aoVal": "D", "content": "$$9{{b}^{2}}+6ab$$ " } ] ]

[ "课内体系->知识点->式->整式的乘除->整式的乘除运算->整式乘除的综合", "课内体系->能力->运算能力" ]

[ "$$18{{b}^{2}}-12ab$$． $${{(2a-3b)}^{2}}-(2a+3b)(2a-3b)$$， $$=4{{a}^{2}}-12ab+9{{b}^{2}}-(4{{a}^{2}}-9{{b}^{2}})$$， $$=4{{a}^{2}}-12ab+9{{b}^{2}}-4{{a}^{2}}+9{{b}^{2}}$$， $$=18{{b}^{2}}-12ab$$． " ]

[ "2016~2017学年10月河南郑州二七区郑州实验外国语中学初一上学期月考第11题3分", "2008年竞赛第7题6分", "2016~2017学年福建泉州南安市南安市实验中学初一下学期期中第16题4分", "2018~2019学年浙江杭州西湖区杭州市之江实验中学初一下学期期末第18题4分" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "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->二元一次方程组的行程问题" ]

[ "设$$18$$路公交车的速度是$$x$$米/分，小王行走的速度是$$y$$米/分，同向行驶的相邻两车的间距为$$s$$米． 每隔$$6$$分钟从背后开过一辆$$18$$路公交车，则 $$6x-6y=s$$．① 每隔$$3$$分钟从迎面驶来一辆$$18$$路公交车，则 $$3x+3y=s$$．② 由①②得，$$s=4x$$，所以$$\\frac{s}{x}=4$$. 即$$18$$路公交车总站发车间隔的时间是$$4$$分钟． " ]

[ "2019~2020学年重庆九龙坡区重庆市育才中学初三上学期期中第11题4分", "2020~2021学年广东深圳福田区深圳市实验学校初二下学期期中第10题3分", "2019~2020学年5月河南郑州中牟县郑州枫杨外国语学校(东校区)初二下学期月考第7题3分", "2020年湖南长沙岳麓区湖南师范大学附属中学初二竞赛（湖南师范大学附属中学教育集团）（6月攀登杯）第6题4分" ]

[ [ { "aoVal": "A", "content": "$$13$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$22$$ " } ] ]

[ "课内体系->能力->运算能力", "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->分式方程的特殊解问题", "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" ]

[ "解不等式组$$\\begin{cases}m-5x\\geqslant 2① x-\\dfrac{11}{2} ~\\textless{} ~3\\left( x+\\dfrac{1}{2} \\right)② \\end{cases}$$， 解不等式①得：$$-5x\\geqslant 2-m$$ $$ x\\leqslant \\frac{m-2}{5}$$， 解不等式②得：$$2x-11 ~\\textless{} ~6x+3$$ $$ 2x-6x ~\\textless{} ~14$$ $$ -4x ~\\textless{} ~14$$ $$ x\\textgreater-\\frac{7}{2}$$， ∴原不等式组的解集为$$-\\frac{7}{2} ~\\textless{} ~x\\leqslant \\frac{m-2}{5}$$， ∵原不等式组有且仅有四个整数解，它们分别为$$-3$$、$$-2$$、$$-1$$，$$0$$， ∴$$0\\leqslant \\frac{m-2}{5} ~\\textless{} ~1$$， ∴$$0\\leqslant m-2 ~\\textless{} ~5$$， ∴$$2\\leqslant m ~\\textless{} ~7$$， 解分式方程$$\\frac{2-my}{2-y}-\\frac{8}{y-2}=1$$， 去分母得$$my-2-8=y-2$$， $$\\left( m-1 \\right)y=8$$ $$y=\\frac{8}{m-1}$$， ∵分式方程有非负数解， ∴$$\\frac{8}{m-1}\\geqslant 0$$且$$\\frac{8}{m-1}\\ne 2$$， ∴$$m\\textgreater1$$且$$m\\ne 5$$， ∵$$m$$为整数， ∴$$m$$可以取$$2$$、$$3$$、$$4$$、$$6$$， 它们的和为$$2+3+4+6=15$$， ∴符合条件的所有整数$$m$$的和为$$15$$， 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->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", "课内体系->能力->运算能力" ]

[ "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． " ]

[ "2001年第18届全国初中数学联赛竞赛第9题" ]

[ [ { "aoVal": "A", "content": "$$43$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$34$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ] ]

[ "知识标签->题型->方程与不等式->一元二次方程->根与系数的关系->题型：韦达定理应用", "知识标签->知识点->方程与不等式->一元二次方程->一元二次方程根的判别式", "知识标签->知识点->方程与不等式->一元二次方程->一元二次方程的根与系数的关系" ]

[ "由于$$xy+\\left( x+y \\right)=23$$，$$xy\\cdot \\left( x+y \\right)=120$$， 则$$x$$，$$y$$与$$\\left( x+y \\right)$$为方程$${{t}^{2}}-23t+120=0$$的两个根，得到$${{t}_{1}}=8$$，$${{t}_{2}}=15$$， 即$$xy=8$$，$$x+y=15$$ ①，或者$$xy=15$$，$$x+y=8$$ ②， ①的时候$$x$$，$$y$$为方程$${{u}^{2}}-15u+8=0$$的根，$${{\\Delta }_{1}}={{15}^{2}}-32=193$$，不是完全平方数，$$xy$$不可能为题目中要求的正整数，舍， ②的时候$$x$$，$$y$$为方程$${{u}^{2}}-8u+15=0$$的根，$${{u}_{1}}=3$$，$${{u}_{2}}=5$$， 故$${{x}^{2}}+{{y}^{2}}={{\\left( x+y \\right)}^{2}}-2xy=34$$． " ]

[ "2017~2018学年广东汕头金平区初二上学期期末第7题3分", "2014~2015学年重庆开县初二上学期期末第7题3分", "2017~2018学年北京朝阳区朝阳外国语学校初二上学期单元测试《数学基础知识竞赛》第18题3分" ]

[ [ { "aoVal": "A", "content": "$$(x-2)(x+9)$$ " } ], [ { "aoVal": "B", "content": "$$(x+2)(x+9)$$ " } ], [ { "aoVal": "C", "content": "$$(x-3)(x+6)$$ " } ], [ { "aoVal": "D", "content": "$$(x-1)(x+18)$$ " } ] ]

[ "课内体系->知识点->式->整式的乘除->整式的乘除运算->多项式乘多项式", "课内体系->能力->运算能力" ]

[ "$$\\text{B}$$、$$(x+2)(x+9)={{x}^{2}}+11x+18$$， $$\\text{C}$$、$$(x-3)(x+6)={{x}^{2}}+3x-18$$， $$\\text{D}$$、$$(x-1)(x+18)={{x}^{2}}+17x-18$$． " ]

[ "2013年第30届全国全国初中数学联赛竞赛第2题7分" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ] ]

[ "课内体系->思想->分类讨论思想", "课内体系->能力->运算能力", "课内体系->知识点->式->整式的乘除->幂的运算->零指数幂", "课内体系->知识点->式->整式的乘除->幂的运算->幂的综合运算" ]

[ "分三种情况进行讨论： ①若$$2-m=1$$，即$$m=1$$时，满足已知等式； ②若$$2-m=-1$$，即$$m=3$$时， $${{\\left( 2-m \\right)}^{{{m}^{2}}-m-2}}={{(-1)}^{4}}=1$$满足已知等式； ③若$$2-m\\ne \\pm 1$$，即$$m\\ne 1$$且$$m\\ne 3$$时， 由已知，得$$\\begin{cases}2-m\\ne 0 {{m}^{2}}-m-2=0 \\end{cases}$$，解得$$m=-1$$， 所以满足等式$${{\\left( 2-m \\right)}^{{{m}^{2}}-m-2}}=1$$的所有实数$$m$$的和$$1+3+(-1)=3$$． " ]

[ "2001年第12届希望杯初一竞赛第2试第5题" ]

[ [ { "aoVal": "A", "content": "$$273$$ " } ], [ { "aoVal": "B", "content": "$$819$$ " } ], [ { "aoVal": "C", "content": "$$1911$$ " } ], [ { "aoVal": "D", "content": "$$3549$$ " } ] ]

[ "竞赛->知识点->数论->整除->因数与倍数" ]

[ "两个正整数为$$a$$与$$b$$， 则$$a+b=60={{2}^{2}}\\times 3\\times 5$$， $$[a,b]=273=3\\times 7\\times 13$$， 显然$$a$$，$$b$$的最大公约数是$$1$$或$$3$$， 如果$$(a,b)=1$$，则$$[a,b]=a\\times b$$， $$a$$，$$b$$只能取$$(21,13)$$，$$(7,39)$$，$$(1,273)$$，$$(3,91)$$，其和均不为$$60$$， 因此$$(a,b)=3$$， 于是$$a=3\\times 7$$，$$b=3\\times 13$$， 所以$$a\\times b=\\left( 3\\times 7 \\right)\\times \\left( 3\\times 13 \\right)=819$$， 故选$$\\text{B}$$． " ]

[ "2018~2019学年9月湖南长沙雨花区湖南广益实验中学初一上学期月考第3题3分", "2019年湖南长沙雨花区湖南广益实验中学初一竞赛（广益杯）第3题3分" ]

[ [ { "aoVal": "A", "content": "不大于$$2$$ " } ], [ { "aoVal": "B", "content": "小于$$2$$ " } ], [ { "aoVal": "C", "content": "不小于$$2$$ " } ], [ { "aoVal": "D", "content": "大于$$2$$ " } ] ]

[ "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" ]

[ "∵$${{x}^{2}}\\geqslant 0$$， ∴$${{x}^{2}}+2\\geqslant 2$$， 即$${{x}^{2}}+2$$的值总是不小于$$2$$． 故选$$\\text{C}$$． " ]

[ "2000年希望杯初二竞赛二试" ]

[ [ { "aoVal": "A", "content": "直角三角形 " } ], [ { "aoVal": "B", "content": "等腰三角形 " } ], [ { "aoVal": "C", "content": "等边三角形 " } ], [ { "aoVal": "D", "content": "直角三角形或等腰三角形 " } ] ]

[ "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->与三角形有关的线段->三角形的三边关系", "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->与三边关系有关的证明" ]

[ "因为$$a$$，$$b$$，$$c$$均为质数且$$a+b+c=16$$，所以$$a$$，$$b$$，$$c$$中有一数为$$2$$，设$$a=2$$，则$$b+c=14$$， 所以$$\\textbar b-c\\textbar\\textless{}2$$．从而有$$\\textbar b-c\\textbar=0$$或$$\\textbar b-c\\textbar=1$$．当$$\\textbar b-c\\textbar=1$$时，$$b$$，$$c$$均不是整数，不合题意．因 此，只有$$\\textbar b-c\\textbar=0$$，即$$a=2$$，$$b=c=7$$，所以三角形是等腰三角形． " ]

[ "1995年第6届全国希望杯初一竞赛复赛第5题", "初一上学期单元测试《有理数》数轴的概念及应用第10题" ]

[ [ { "aoVal": "A", "content": "$$1994$$或$$1995$$ " } ], [ { "aoVal": "B", "content": "$$1994$$或$$1996$$ " } ], [ { "aoVal": "C", "content": "$$1995$$或$$1996$$ " } ], [ { "aoVal": "D", "content": "$$1995$$或$$1997$$ " } ] ]

[ "课内体系->能力->空间想象能力", "课内体系->能力->运算能力", "课内体系->知识点->数->有理数->数轴->数轴上的规律探究" ]

[ "若所画的长为$$1995$$厘米的线段的两个端点$$A$$与$$B$$均为整点时， 此时线段$$AB$$盖住的整点个数是$$1995+1=1996$$个． 若$$A$$点不是整点，则$$B$$点也不是整点，此时线段$$AB$$盖住的整点个数为$$1995$$个， 所以长为$$1995$$厘米的线段盖住的整点是$$1995$$个， 所以长为$$1995$$厘米的线段盖住的整点是$$1995$$或$$1996$$个．选$$\\text{C}$$． " ]

[ "2014年第25届全国希望杯初一竞赛初赛第10题4分" ]

[ [ { "aoVal": "A", "content": "$$14$$平方厘米 " } ], [ { "aoVal": "B", "content": "$$103$$平方厘米 " } ], [ { "aoVal": "C", "content": "$$214$$平方厘米 " } ], [ { "aoVal": "D", "content": "$$400$$平方厘米 " } ] ]

[ "课内体系->思想->方程思想", "课内体系->知识点->式->整式的乘除->整式乘除化简求值->整式乘除化简求值-整体代入化简求值", "课内体系->能力->运算能力", "课内体系->能力->推理论证能力" ]

[ "设长方体的长、宽、高分别为$$a$$厘米，$$b$$厘米，$$c$$厘米， 则长方体的表面积为$$2(ab+bc+ac)$$平方厘米， 将长方体的长、宽、高都增加$$1$$厘米后， 长方体的表面积增加的值如下： $$2[(a+1)(b+1)+(b+1)(c+1)+(a+1)(c+1)]-2(ab+bc+ac)$$ $$=2(a+b+1+b+c+1+a+c+1)$$ $$=4(a+b+c)+6$$． ∵$$a$$，$$b$$，$$c$$都是正整数， ∴$$4(a+b+c)+6\\geqslant 4(1+1+1)+6=18$$，且能写成$$4m+6$$的形式，其中$$m$$为正整数， 选项中所给数字，只有$$214$$可以表示为$$4\\times 52+6$$的形式，故选$$\\text{C}$$． " ]

[ "2018~2019学年甘肃兰州城关区兰州市第三十五中学初二下学期期中第12题4分", "初一下学期单元测试《不等式与不等式组》一元一次不等式及其应用第28题", "2011年第16届华杯赛初一竞赛初赛第3题" ]

[ [ { "aoVal": "A", "content": "$$x\\textgreater\\frac{1}{5}$$ " } ], [ { "aoVal": "B", "content": "$$x\\textless{}-\\frac{1}{5}$$ " } ], [ { "aoVal": "C", "content": "$$x\\textgreater-\\frac{1}{5}$$ " } ], [ { "aoVal": "D", "content": "$$x\\textless{}\\frac{1}{5}$$ " } ] ]

[ "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" ]

[ "方法一：原不等式变形得：$$\\frac{x}{a}\\textgreater-\\frac{1}{b}$$， ∵$$x\\textless{}\\frac{1}{5}$$， ∴$$a\\textless{}0$$． 解不等式得：$$x\\textless{}-\\frac{a}{b}$$，$$-\\frac{a}{b}=\\frac{1}{5}$$，即$$b=-5a$$． ∴$$-5ax-a\\textgreater0$$， ∴$$x\\textgreater-\\frac{1}{5}$$． 方法二：因为不等式等$$\\frac{x}{a}+\\frac{1}{b}\\textgreater0$$的解集为$$x ~\\textless{} ~\\frac{1}{5}$$，因此必有$$a ~\\textless{} ~0$$，所以$$\\frac{-a}{b}=\\frac{1}{5}$$，并且$$b\\textgreater0$$，所以由$$bx-a\\textgreater0$$得到$$x\\textgreater-\\frac{1}{5}$$． " ]

[ "2017年湖南长沙天心区湘郡培粹实验中学初二竞赛（觉园杯）第5题4分", "2019年湖南长沙天心区湘郡培粹实验中学初二竞赛初赛(觉园杯)第5题4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->思想->整体思想", "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->能力->运算能力", "课内体系->方法->整体法" ]

[ "∵$$abc \\textless{} 0$$， ∴$$a$$，$$b$$，$$c$$中三个都是负数或两正数，一个是负数，当三个都是负数时：若``$$\\frac{x-y}{a}=abc$$，则$$x-y=a^{2}bc\\textgreater0$$，即$$x \\textgreater y$$，同理可得：$$y \\textgreater z$$，$$z \\textgreater x$$这三个式子不能同时成立，即$$a$$，$$b$$，$$c$$不能同时是负数．则$$P\\left( ab,bc \\right)$$不可能在第一象限． 所以选$$\\text{A}$$． " ]

[ "1985年全美数学竞赛（AMC）竞赛第10题" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "课内体系->知识点->数->有理数->有理数的概念及分类->有理数的定义", "美国AMC8->Knowledge Point->Number and Operations->Decimals->Recurring Decimal" ]

[ "$$\\frac{1}{7}= 0.142857$$$$142857142857$$$$\\cdots $$，所以有$$6$$位重复的小数 ，所以第$$1985$$位是$$5$$. 故选$$\\text{C}$$． " ]

[ "1992年第9届全国初中数学联赛竞赛第6题" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ] ]

[ "课内体系->知识点->四边形->特殊平行四边形->正方形->正方形的性质" ]

[ "据正方形的对称性，只需考虑它的$$\\frac{1}{4}$$部分即可，记圆周经过的所有小方格的圆内部分的$$4$$面积之和为$$S_{1}^{\\^{\\prime }}$$，圆周经过的所有小方格的圆外部分的面积之和为$$S_{2}^{\\^{\\prime }}$$， 则$$S_{1}^{\\^{\\prime }}=4\\pi -8$$，$$S_{2}^{\\^{\\prime }}=15-4\\pi $$， $$\\therefore \\frac{S_{1}^{\\^{\\prime }}}{S_{2}^{\\^{\\prime }}}=\\frac{4{{S}_{1}}}{4{{S}_{2}}}=\\frac{4\\pi -8}{15-4\\pi }\\approx \\frac{4.56}{2.44}$$ 故$$\\frac{{{S}_{1}}}{{{S}_{2}}}$$的整数部分是$$1$$ . 故选$$\\text{B}$$ . " ]

[ "1999年第10届希望杯初二竞赛第2试第4题" ]

[ [ { "aoVal": "A", "content": "$$x\\textgreater y$$ " } ], [ { "aoVal": "B", "content": "$$x \\textless{} y$$ " } ], [ { "aoVal": "C", "content": "$$x-y$$ " } ], [ { "aoVal": "D", "content": "随$$a$$，$$b$$，$$c$$的取值而变化 " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->式->分式->分式的运算->分式加减、乘除、乘方混合运算" ]

[ "由题意有$$2x-2y=\\frac{2}{a}+\\frac{2}{b}+\\frac{2}{c}-\\frac{2}{\\sqrt{ab}}-\\frac{2}{\\sqrt{bc}}-\\frac{2}{\\sqrt{ac}}$$ $$=(\\frac{1}{a}-\\frac{2}{\\sqrt{ab}}+\\frac{1}{b})+(\\frac{1}{b}-\\frac{2}{\\sqrt{bc}}+\\frac{1}{c})+(\\frac{1}{c}-\\frac{2}{\\sqrt{ac}}-+\\frac{1}{a})$$ $$={{(\\frac{1}{\\sqrt{a}}-\\frac{1}{\\sqrt{b}})}^{2}}+{{(\\frac{1}{\\sqrt{b}}-\\frac{1}{\\sqrt{c}})}^{2}}+{{(\\frac{1}{\\sqrt{c}}-\\frac{1}{\\sqrt{a}})}^{2}}$$ 又∵$$a\\ne b$$， ∴$${{(\\frac{1}{\\sqrt{a}}-\\frac{1}{\\sqrt{b}})}^{2}}\\textgreater0$$，$${{(\\frac{1}{\\sqrt{b}}-\\frac{1}{\\sqrt{c}})}^{2}}\\geqslant 0$$，$${{(\\frac{1}{\\sqrt{c}}-\\frac{1}{\\sqrt{a}})}^{2}}\\geqslant 0$$． ∴$$x-y\\textgreater0$$，$$x\\textgreater y$$．选$$\\text A$$． " ]

[ "2013年第24届全国希望杯初二竞赛复赛第7题4分" ]

[ [ { "aoVal": "A", "content": "冠军是甲 " } ], [ { "aoVal": "B", "content": "冠军是乙 " } ], [ { "aoVal": "C", "content": "冠军是丙 " } ], [ { "aoVal": "D", "content": "甲、乙、丙同时遇到丁 " } ] ]

[ "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->一元一次方程的行程问题->一元一次方程的行程问题-变速问题" ]

[ "出发时间$$t$$后，甲、乙、丙与丁的距离为$${{s}_{i}}=({{v}_{i}}-{{v}_{4}})t$$，$$i=1$$，$$2$$，$$3$$． 甲、乙、丙反向后，遇到丁的时间为$${{t}_{i}}^{\\prime }=\\frac{({{v}_{i}}-{{v}_{4}})t}{{{v}_{i}}+{{v}_{4}}}=\\left( 1-\\frac{2{{v}_{4}}}{{{v}_{i}}+{{v}_{4}}} \\right)t$$． 已知$${{v}_{1}}\\textgreater{{v}_{2}}\\textgreater{{v}_{3}}\\textgreater{{v}_{4}}\\textgreater0$$， 故有$${{t}_{1}}\\textgreater{{t}_{2}}\\textgreater{{t}_{3}}$$， 即丙先遇到丁，丙是冠军． " ]

[ "2015~2016学年浙江温州乐清市乐清市育英寄宿学校初二上学期期中实验a班第4题4分", "2019~2020学年12月浙江杭州上城区北京师范大学附属杭州中学初三上学期周测A卷第8题3分", "初三上学期单元测试《概率初步》第23题", "2008年竞赛第2题6分", "2020~2021学年10月浙江杭州下城区杭州市景成实验学校中学部初三上学期月考第8题3分", "2016~2017学年9月浙江杭州拱墅区杭州锦绣·育才中学附属学校初三上学期月考第6题3分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{5}{12}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{4}{9}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{17}{36}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{2}$$ " } ] ]

[ "课内体系->知识点->统计与概率->概率->概率的计算方法->列举法求概率" ]

[ "掷骰子有$$6\\times 6=36$$种情况． 根据题意有：$$4n-{{m}^{2}}\\textless{}0$$， 因此满足的点有：$$n=1$$，$$m=3$$，$$4$$，$$5$$，$$6$$， $$n=2$$，$$m=3$$，$$4$$，$$5$$，$$6$$， $$n=3$$，$$m=4$$，$$5$$，$$6$$， $$n=4$$，$$m=5$$，$$6$$， $$n=5$$，$$m=5$$，$$6$$， $$n=6$$，$$m=5$$，$$6$$， 共有$$17$$种， 故概率为：$$17\\div 36=\\frac{17}{36}$$． " ]

[ "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->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", "课内体系->能力->运算能力" ]

[ "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． " ]

[ "2018~2019学年甘肃兰州城关区兰州市第三十五中学初二下学期期中第12题4分", "2011年第16届华杯赛初一竞赛初赛第3题", "初一下学期单元测试《不等式与不等式组》一元一次不等式及其应用第28题" ]

[ [ { "aoVal": "A", "content": "$$x\\textgreater\\frac{1}{5}$$ " } ], [ { "aoVal": "B", "content": "$$x\\textless{}-\\frac{1}{5}$$ " } ], [ { "aoVal": "C", "content": "$$x\\textgreater-\\frac{1}{5}$$ " } ], [ { "aoVal": "D", "content": "$$x\\textless{}\\frac{1}{5}$$ " } ] ]

[ "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）->含参不等式->由一个已知不等式的解集求另一个不等式的解集" ]

[ "不等式$$ \\frac{x}{a}+ \\frac{1}{b}\\textgreater0 $$的解集为$$\\frac{x}{a}\\textgreater- \\frac{1}{b}$$， $$x\\textless{}- \\frac{a}{b}$$， $$x\\textless{} \\frac{1}{5}$$， 所以$$ \\frac{a}{b}=- \\frac{1}{5} $$且$$a\\textless{}0$$，$$b\\textgreater0$$， 所以不等式$$bx-a\\textgreater0$$的解集为$$bx\\textgreater a$$， $$x\\textgreater{} \\frac{a}{b}$$， $$x\\textgreater- \\frac{1}{5}$$， 故选：$$\\text{C}$$． " ]

[ "2013年第24届全国希望杯初二竞赛初赛第2题4分" ]

[ [ { "aoVal": "A", "content": "$$m\\textgreater n$$ " } ], [ { "aoVal": "B", "content": "$$m\\textless{}n$$ " } ], [ { "aoVal": "C", "content": "$$m=n$$ " } ], [ { "aoVal": "D", "content": "不能确定的 " } ] ]

[ "课内体系->知识点->函数->一次函数->一次函数基础", "课内体系->能力->运算能力" ]

[ "因为直线$$y=-2x-6$$的斜率$$k=-2\\textless{}0$$， 所以$$y$$随$$x$$的增大而减小． 由于$$-7\\textgreater-8$$， 所以$$m\\textless{}n$$． " ]

[ "2008年第19届希望杯初二竞赛第1试第3题" ]

[ [ { "aoVal": "A", "content": "$$\\sqrt{a}$$，$$\\sqrt{b}$$都是有理数 " } ], [ { "aoVal": "B", "content": "$$\\sqrt{a}$$，$$\\sqrt{b}$$都是无理数 " } ], [ { "aoVal": "C", "content": "$$\\sqrt{a}$$，$$\\sqrt{b}$$都是有理数或都是无理数 " } ], [ { "aoVal": "D", "content": "$$\\sqrt{a}$$，$$\\sqrt{b}$$中有理数和无理数各有一个 " } ] ]

[ "课内体系->知识点->数->实数->无理数有关的计算->无理数的定义" ]

[ "$$a$$，$$b$$都是有理数， ∴$$a-b$$是有理数． 又$$\\sqrt{a}+\\sqrt{b}$$是有理数， ∴$$\\sqrt{a}-\\sqrt{b}=\\frac{a-b}{\\sqrt{a}+\\sqrt{b}}$$是有理数． 于是$$\\sqrt{a}=\\frac{1}{2}\\left[ (\\sqrt{a}+\\sqrt{b})+(\\sqrt{a}-\\sqrt{6}) \\right]$$是有理数 ，$$\\sqrt{b}=\\frac{1}{2}\\left[ (\\sqrt{a}+\\sqrt{b})-(\\sqrt{a}-\\sqrt{b}) \\right]$$是有理数． 故选$$\\text{A}$$． " ]

[ "2008年第19届希望杯初二竞赛第1试第11题" ]

[ [ { "aoVal": "A", "content": "$$243$$ " } ], [ { "aoVal": "B", "content": "$$252$$ " } ], [ { "aoVal": "C", "content": "$$257$$ " } ], [ { "aoVal": "D", "content": "$$506$$ " } ], [ { "aoVal": "E", "content": "$$2010$$ " } ] ]

[ "课内体系->知识点->数->有理数->数的特征->因数和倍数", "竞赛->知识点->数论->整除->分解素因数" ]

[ "$$2008$$分解成$$2\\times 4\\times 251$$时， $$a+b+c=2+4+251=257$$． 这时$$a+b+c$$的值最小． " ]

[ "2012年第23届全国希望杯初一竞赛初赛第2题4分" ]

[ [ { "aoVal": "A", "content": "$$508.5$$ " } ], [ { "aoVal": "B", "content": "$$511.3$$ " } ], [ { "aoVal": "C", "content": "$$462.8$$ " } ], [ { "aoVal": "D", "content": "$$605.5$$ " } ] ]

[ "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->能力->推理论证能力", "课内体系->知识点->数->有理数->有理数与实际问题->有理数加减法与实际问题" ]

[ "由景山的资料知道，北京地平面的海拔高度是$$94.2-45.7=48.5$$（米）， 而$$557-48.5=508.5$$（米），这就是香炉峰的相对高度． " ]

[ "2019年湖南长沙雨花区湖南广益实验中学初一竞赛（广益杯）第5题3分" ]

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

[ "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-利用数轴" ]

[ "有无数个有理数． " ]

[ "2012年第23届全国希望杯初二竞赛初赛第10题4分" ]

[ [ { "aoVal": "A", "content": "$$2225$$ " } ], [ { "aoVal": "B", "content": "$$2226$$ " } ], [ { "aoVal": "C", "content": "$$2227$$ " } ], [ { "aoVal": "D", "content": "$$2228$$ " } ] ]

[ "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->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" ]

[ "因为$${{x}_{1}}$$，$${{x}_{2}}$$，$$\\cdots $$，$${{x}_{100}}$$是自然数，且$${{x}_{1}}\\textless{}{{x}_{2}}\\textless{}\\cdots \\textless{}{{x}_{100}}$$， 所以$$7001={{x}_{1}}+{{x}_{2}}+\\cdots +{{x}_{100}}\\geqslant {{x}_{1}}+({{x}_{1}}+1)+({{x}_{1}}+2)+\\cdots +({{x}_{1}}+99)$$， 即$$7001\\geqslant 100{{x}_{1}}+4950$$， 解得$${{x}_{1}}\\leqslant 20$$． 又因为$$20+21+22+\\cdots +119=6950=7001-51$$， $$21+22+23+\\cdots +120=7050\\textgreater7001$$， 所以，当$${{x}_{1}}=20$$，$${{x}_{2}}=21$$，$$\\cdots $$，$${{x}_{49}}=68$$，$${{x}_{50}}=70$$，$${{x}_{51}}=71$$，$$\\cdots $$，$${{x}_{100}}=120$$时， $${{x}_{1}}+{{x}_{2}}+\\cdots +{{x}_{50}}$$取得最大值$$2226$$． " ]

[ "2011年第22届全国希望杯初一竞赛初赛第10题4分" ]

[ [ { "aoVal": "A", "content": "$$25$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ] ]

[ "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->一元一次方程的和差倍分" ]

[ "甲每分钟走全程的$$\\frac{1}{30}$$，乙每分钟走全程的$$\\frac{1}{40}$$， 设甲$$t$$分钟追上乙， 则$$\\frac{t}{30}=\\frac{(6+t)}{40}$$，解得$$t=18$$（分钟）． $$40-6-18=16$$（分钟）． 此后，乙还要走$$16$$分钟才能到达$$B$$． " ]

[ "1996年第13届全国初中数学联赛竞赛第6题" ]

[ [ { "aoVal": "A", "content": "$$4$$个 " } ], [ { "aoVal": "B", "content": "$$8$$个 " } ], [ { "aoVal": "C", "content": "$$12$$个 " } ], [ { "aoVal": "D", "content": "$$24$$个 " } ] ]

[ "课内体系->知识点->圆->圆与多边形->正多边形与圆->圆内接正多边形相关计算" ]

[ "$$\\text{C}$$：$$12$$个；$$\\frac{20}{4}+\\frac{20}{5}+\\frac{20}{10}+\\frac{20}{20}=12$$． ", "<p>设正$$k$$边形满足条件，则除去$$k$$个顶点外的$$20-k$$个点均匀的分布在正$$k$$边形各边所对的劣弧上，则有：</p>\n<p>于是$$\\frac{20-k}{k}=\\frac{20}{k}-1$$ 是整数，故$$k$$能整除$$20$$，同时$$k\\geqslant 3$$3．</p>\n<p>所以$$k=4$$或$$5$$或$$10$$或$$20$$，</p>\n<p>故所求正多边形的个数为$$\\frac{20}{4}+\\frac{20}{5}+\\frac{20}{10}+\\frac{20}{20}=12$$．</p>\n<p>故选$$C$$．</p>" ]

[ "1996年第7届全国希望杯初一竞赛复赛第8题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{4}$$，$$\\frac{1}{6}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1}{4}$$，$$\\frac{1}{12}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{6}$$，$$\\frac{1}{10}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{1}{10}$$，$$\\frac{1}{8}$$ " } ] ]

[ "课内体系->知识点->数->有理数->有理数基础运算->有理数加法->有理数加法运算", "课内体系->能力->运算能力" ]

[ "易知$$\\frac{1}{4}+\\frac{1}{6}+\\frac{1}{12}=\\frac{1}{2}$$，所以$$\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{6}+\\frac{1}{12}=1$$， 因此删去的两个加数为$$\\frac{1}{8}$$和$$\\frac{1}{10}$$，选$$\\text{D}$$． " ]

[ "1992年第3届全国希望杯初一竞赛复赛第4题" ]

[ [ { "aoVal": "A", "content": "奇数 " } ], [ { "aoVal": "B", "content": "偶数 " } ], [ { "aoVal": "C", "content": "负整数 " } ], [ { "aoVal": "D", "content": "非负整数 " } ] ]

[ "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->有理数减法->有理数加减混合运算" ]

[ "由于两个整数$$a$$，$$b$$前面任意添加``$$+$$''号或``$$-$$''号，其代数和的奇偶性不变． 这个性质对$$n$$个整数也是正确的． 因此，$$1$$，$$2$$，$$3$$，$$\\cdots $$，$$1991$$，$$1992$$的每一个数前面任意添上``$$+$$''号或``$$-$$''号，其代数和的奇偶性与$$(-1)+2-3+4-5+6-7+8-\\cdots -1991+1992=996$$的奇偶性相同，是偶数，所以选$$\\text{B}$$． " ]

[ "1999年第10届希望杯初一竞赛第1题" ]

[ [ { "aoVal": "A", "content": "$$1999$$ " } ], [ { "aoVal": "B", "content": "$$-1999$$ " } ], [ { "aoVal": "C", "content": "$$-\\frac{1}{1999}$$ " } ], [ { "aoVal": "D", "content": "$$\\left\\textbar{} -\\frac{1}{1999} \\right\\textbar$$ " } ] ]

[ "课内体系->知识点->数->有理数->相反数->相反数的性质" ]

[ "根据相反数的定义，$$\\frac{1}{1999}$$的相反数是$$-\\frac{1}{1999}$$． 故选$$\\text{C}$$． " ]

[ "1998年全美数学竞赛（AMC）竞赛" ]

[ [ { "aoVal": "A", "content": "$$\\frac 6x$$ " } ], [ { "aoVal": "B", "content": "$$\\frac 6{x+1}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac 6{x-1}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac x6$$ " } ], [ { "aoVal": "E", "content": "$$\\frac {x+1}6$$ " } ] ]

[ "美国AMC8->Knowledge Point->Number and Operations->Fractions->Fraction Operations", "课内体系->知识点->数->有理数->分数" ]

[ "设$$x=7$$，下面哪个最小？ 最小的分数形式是$$b$$，其中$$b$$大于$$a$$． 在这个问题中，我们需要所有给定值中的最大可能值以分母表示．这个值是$$x+1$$或8 分子越小，就是$$6$$． 带有$$\\frac {6}{x+1}$$的答案选项是$$\\rm B$$． 为每个答案选项代入$$x$$会给出 ($$\\rm A$$)$$\\frac 67$$ ($$\\rm B$$)$$\\frac 68$$ ($$\\rm C$$)$$\\frac 66$$ ($$\\rm D$$)$$\\frac 76$$ ($$\\rm E$$)$$\\frac 86$$ 从这里，我们可以看到最小的是答案选项$$\\rm B$$． The smallest fraction would be in the form $$\\frac{a}{b}$$ where $$b$$ is larger than $$a$$. In this problem, we would need the largest possible value out of all the given values to be in the denominator. This value is $$x+1$$ or $$8$$. The smaller would go on the numerator, which is $$6$$. The answer choice with $$\\frac{6}{x+1}$$ is $$\\text{B}$$. Plugging $$x$$ in for every answer choice would give ($$\\rm A$$)$$\\frac 67$$ ($$\\rm B$$)$$\\frac 68$$ ($$\\rm C$$)$$\\frac 66$$ ($$\\rm D$$)$$\\frac 76$$ ($$\\rm E$$)$$\\frac 86$$ From here, we can see that the smallest is answer choice $$\\text{B}$$. " ]

[ "1996年第7届希望杯初二竞赛第1题4分" ]

[ [ { "aoVal": "A", "content": "$${{y}^{2}}-{{x}^{2}}$$ " } ], [ { "aoVal": "B", "content": "$${{x}^{2}}-{{y}^{2}}$$ " } ], [ { "aoVal": "C", "content": "$${{x}^{2}}-4{{y}^{2}}$$ " } ], [ { "aoVal": "D", "content": "$$4{{x}^{2}}-{{y}^{2}}$$ " } ] ]

[ "竞赛->知识点->数与式->分式->分式的基本运算", "课内体系->知识点->式->分式->分式的基础->分式的性质", "课内体系->知识点->式->分式->分式的运算->分式约分" ]

[ "原式$$=\\left( x-y+\\frac{4xy}{x-y} \\right)\\left( x+y-\\frac{4xy}{x+y} \\right)$$ $$=\\frac{{{x}^{2}}-2xy+{{y}^{2}}+4xy}{x-y}\\cdot \\frac{{{x}^{2}}+2xy+{{y}^{2}}-4xy}{x+y}$$ $$=\\frac{{{(x+y)}^{2}}}{x-y}\\cdot \\frac{{{(x-y)}^{2}}}{x+y}$$ $$=(x+y)(x-y)$$ $$={{x}^{2}}-{{y}^{2}}$$． 故选$$\\text{B}$$． " ]

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

[ [ { "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" ]

[ "\\textbf{（知识点：二次根式化简求值）} ∵$$x=\\frac{1+\\sqrt{1994}}{2}$$， ∴$${{(2x-1)}^{2}}=1994$$，即$$4{{x}^{2}}-4x-1993=0$$， ∴$${{(4{{x}^{3}}-1997x-1994)}^{2001}}$$ $$={{[(4{{x}^{2}}-4x-1993)x+(4{{x}^{2}}-4x-1993)-1]}^{2001}}$$ $$={{(-1)}^{2001}}$$ $$=-1$$． " ]

[ "2002年第13届希望杯初一竞赛第1试第2题" ]

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

[ "竞赛->知识点->数与式->数的运算->有理数运算问题" ]

[ "由相反数定义可知，$$\\frac{1}{a}$$的相反数是$$-\\frac{1}{a}$$， 故选$$\\text{C}$$． " ]

[ "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": "$$20$$ " } ] ]

[ "知识标签->学习能力->运算能力", "知识标签->题型->式->整式的乘除->乘法公式->题型：利用完全平方公式计算", "知识标签->知识点->式->分式->分式的运算->分式的加减", "知识标签->知识点->式->整式的乘除->乘法公式->完全平方公式" ]

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

[ "2014年第25届全国希望杯初一竞赛初赛第8题4分" ]

[ [ { "aoVal": "A", "content": "$$A-B$$一定是多项式 " } ], [ { "aoVal": "B", "content": "$$A-B$$是次数不低于$$6$$的整式 " } ], [ { "aoVal": "C", "content": "$$A+B$$一定是单项式 " } ], [ { "aoVal": "D", "content": "$$A+B$$是次数不高于$$6$$的整式 " } ] ]

[ "课内体系->知识点->式->整式的加减->整式有关的概念->整式->整式的定义", "课内体系->知识点->式->整式的加减->整式有关的概念->多项式->多项式的项" ]

[ "若$$A=B$$，则$$A-B=0$$，为单项式，不是多项式，次数为$$0$$； 若$$A=2{{a}^{6}}+{{a}^{5}}$$，$$B=3{{a}^{6}}$$，则$$A+B=5{{a}^{6}}+{{a}^{5}}$$，为多项式． 由上可知，$$\\text{D}$$选项正确． " ]

[ "2018~2019学年5月天津南开区天津市南开中学初一下学期月考第12题3分", "2000年竞赛（全国初中数学竞赛）第1题5分" ]

[ [ { "aoVal": "A", "content": "$$M=P$$ " } ], [ { "aoVal": "B", "content": "$$M\\textgreater P$$ " } ], [ { "aoVal": "C", "content": "$$M \\textless{} P$$ " } ], [ { "aoVal": "D", "content": "不确定 " } ] ]

[ "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-作差法", "课内体系->能力->运算能力" ]

[ "由题意得：$$a+b+c=3M$$， $$a+b=2N$$，$$N+c=2P$$， ∴$$M=\\frac{a+b+c}{3}$$， $$P=\\frac{N+c}{2}$$， $$N=\\frac{a+b}{2}$$， ∴将$$N$$代入$$P$$可得： $$P=\\frac{a+b+2c}{4}$$， $$M-P=\\frac{a+b-2c}{12}$$， 又∵$$a\\textgreater b\\textgreater c$$， ∴$$a++c\\textgreater3c$$， ∴$$M-P\\textgreater0$$， ∴$$M\\textgreater P$$． 故选$$\\text{B}$$． " ]

[ "2010年第21届全国希望杯初一竞赛初赛第6题4分" ]

[ [ { "aoVal": "A", "content": "$$54{}^{}\\circ $$ " } ], [ { "aoVal": "B", "content": "$$81{}^{}\\circ $$ " } ], [ { "aoVal": "C", "content": "$$99{}^{}\\circ $$ " } ], [ { "aoVal": "D", "content": "$$162{}^{}\\circ $$ " } ] ]

[ "知识标签->题型->几何图形初步->相交线与平行线->相交线->题型：计算对顶角、邻补角度数", "知识标签->知识点->几何图形初步->相交线与平行线->相交线->邻补角", "知识标签->知识点->几何图形初步->角->角的和差" ]

[ "∵$$\\angle AOB$$和$$\\angle BOC$$互为邻补角， ∴$$\\angle AOB+\\angle BOC=180{}^{}\\circ $$． ∵$$\\angle AOB$$比$$\\angle BOC$$大$$18{}^{}\\circ $$， ∴$$\\angle AOB-\\angle BOC=18{}^{}\\circ $$． 两式相加，化简得$$\\angle AOB=99{}^{}\\circ $$． " ]

[ "1993年第4届全国希望杯初一竞赛初赛第7题" ]

[ [ { "aoVal": "A", "content": "$$-a$$是负数 " } ], [ { "aoVal": "B", "content": "$${{a}^{2}}$$是正数 " } ], [ { "aoVal": "C", "content": "$$-\\left\\textbar{} {{a}^{2}} \\right\\textbar$$是负数 " } ], [ { "aoVal": "D", "content": "$${{(a-1993)}^{2}}+0.001$$是正数 " } ] ]

[ "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" ]

[ "当$$a=0$$，显然$$\\text{A}$$，$$\\text{B}$$，$$\\text{C}$$，均不正确，应排除，所以选$$\\text{D}$$． 事实上，对任意有理数$$a$$，都有$${{(a-1993)}^{2}}\\geqslant 0$$， 所以$${{(a-1993)}^{2}}+0.001\\textgreater0$$是正数． " ]

[ "2013年第24届全国希望杯初一竞赛复赛第7题4分" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ] ]

[ "知识标签->知识点->方程与不等式->一元一次方程->解一元一次方程", "知识标签->学习能力->运算能力", "知识标签->题型->方程与不等式->一元一次方程->解一元一次方程->题型：含绝对值的一元一次方程" ]

[ "由$$\\left\\textbar{} x+1 \\right\\textbar$$依次取$$0$$，$$\\pm1$$， 可得方程组$$\\begin{cases}x+1=0 2x-1=\\pm 1 \\end{cases}$$或$$\\begin{cases}x+1=\\pm1 2x-1=0 \\end{cases}$$， 解之，可得两方程均无解． 所以，原方程整数解的个数为$$0$$ " ]

[ "2018~2019学年9月福建福州鼓楼区福州励志中学初三上学期月考第10题4分", "2017年浙江宁波鄞州区鄞州中学自主招生三位一体第8题5分", "2019~2020学年11月浙江杭州江干区杭州市采荷中学初三上学期月考（杭州采荷中学教育集团）第10题3分", "2006年竞赛第3题6分" ]

[ [ { "aoVal": "A", "content": "$$h \\textless{} 1$$ " } ], [ { "aoVal": "B", "content": "$$h=1$$ " } ], [ { "aoVal": "C", "content": "$$1 \\textless{} h \\textless{} 2$$ " } ], [ { "aoVal": "D", "content": "$$h\\textgreater2$$ " } ] ]

[ "课内体系->知识点->函数->二次函数->二次函数与几何综合->二次函数与直角三角形结合", "课内体系->能力->运算能力", "课内体系->能力->推理论证能力" ]

[ "设点$$A$$的坐标为$$\\left( a,{{a}^{2}} \\right)$$，点$$C$$的坐标为$$\\left( c,{{c}^{2}} \\right)$$，$$\\left( \\left\\textbar{} c \\right\\textbar{} \\textless{} \\left\\textbar{} a \\right\\textbar{} \\right)$$，则点$$B$$的坐标为$$\\left( -a,{{a}^{2}} \\right)$$，由勾股定理，得$$A{{C}^{2}}={{\\left( c-a \\right)}^{2}}+{{\\left( {{c}^{2}}-{{a}^{2}} \\right)}^{2}}$$，$$B{{C}^{2}}={{\\left( c+a \\right)}^{2}}+{{\\left( {{c}^{2}}+{{a}^{2}} \\right)}^{2}}$$，$$A{{C}^{2}}+B{{C}^{2}}=A{{B}^{2}}$$，所以$${{\\left( {{a}^{2}}-{{c}^{2}} \\right)}^{2}}={{a}^{2}}-{{c}^{2}}$$．由于$${{a}^{2}}\\textgreater{{c}^{2}}$$，所以$${{a}^{2}}-{{c}^{2}}=1$$，根据图象知斜边$$AB$$上高$${{h}^{2}}={{a}^{2}}-{{c}^{2}}=1$$，所以$$h=1$$，故选$$\\text{B}$$． " ]

[ "2015年第26届全国希望杯初二竞赛复赛第10题" ]

[ [ { "aoVal": "A", "content": "$$1616$$ " } ], [ { "aoVal": "B", "content": "$$1689$$ " } ], [ { "aoVal": "C", "content": "$$2689$$ " } ], [ { "aoVal": "D", "content": "$$2616$$ " } ] ]

[ "课内体系->能力->推理论证能力", "课内体系->知识点->几何图形初步->命题与证明" ]

[ "假设可以使得每$$4$$个箱子里的球个数完全相同，那么此时最少有$$3\\times (0+1+2+\\cdots +32)+33=1617$$个球，于是所求的$$n$$的最大值为$$1616$$． " ]

[ "1992年第3届希望杯初二竞赛第8题" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ] ]

[ "竞赛->知识点->数与式->整式->综合除法和余数定理 " ]

[ "设$$a=7k+4$$（$$k$$为正整数），则 $${{a}^{3}}+5={{\\left( 7k+4 \\right)}^{3}}+5$$ $$={{\\left( 7k \\right)}^{3}}+3\\times {{\\left( 7k \\right)}^{2}}\\times 4+3\\times \\left( 7k \\right)\\times {{4}^{2}}+{{4}^{3}}+5$$ $$=7\\left( {{7}^{2}}{{k}^{3}}+3\\times 7{{k}^{2}}\\times 4+3k\\times {{4}^{2}}+9 \\right)+6$$， 因此，$${{a}^{3}}+5$$被$$7$$除余$$6$$． 故选$$\\text{D}$$． " ]

[ "2010年第15届华杯赛初一竞赛初赛第4题" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$\\pm \\frac{1}{4}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{3}{4}$$ " } ], [ { "aoVal": "D", "content": "$$\\pm \\frac{3}{4}$$ " } ] ]

[ "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->含绝对值的一元一次方程", "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->绝对值的非负性", "竞赛->知识点->数与式->绝对值->给定范围绝对值化简" ]

[ "当$$x\\leqslant 0$$，原方程为：$$\\left\\textbar{} (-x+1)-(-x) \\right\\textbar-(-x+1)+(-x)=1$$， 即$$1+x-1-x=1$$，无解； 当$$0 ~\\textless{} ~x\\leqslant 1$$，原方程为：$$\\left\\textbar{} (1-x)-x \\right\\textbar-(1-x)+x=1$$， 即$$\\left\\textbar{} 1-2x \\right\\textbar-1+2x=1$$，这种情况下当$$0 ~\\textless{} ~x\\leqslant \\frac{1}{2}$$ 时，$$1-2x-1+2x=1$$无解； 当$$\\frac{1}{2} ~\\textless{} ~x\\leqslant 1$$，$$2x-1-1+2x=1$$，$$x=\\frac{3}{4}$$； 当$$x ~\\textless{} ~1$$，原方程为：$$\\left\\textbar{} (x-1)-x \\right\\textbar-(x-1)+x=1$$， 即$$1-(x-1)+x=1$$，无解． 故选$$\\text{C}$$． " ]

[ "2008年第19届希望杯初一竞赛第2试第6题4分" ]

[ [ { "aoVal": "A", "content": "$$2$$组 " } ], [ { "aoVal": "B", "content": "$$4$$组 " } ], [ { "aoVal": "C", "content": "$$6$$组 " } ], [ { "aoVal": "D", "content": "$$8$$组 " } ] ]

[ "竞赛->知识点->方程与不等式->二次方程->二元二次方程组" ]

[ "原方程变形为$$\\left( \\left\\textbar{} x \\right\\textbar+1 \\right)\\left( \\left\\textbar{} y \\right\\textbar-2 \\right)=2$$， 因为$$\\left\\textbar{} x \\right\\textbar+1\\geqslant 1$$， 所以$$\\begin{cases}\\left\\textbar{} x \\right\\textbar+1=1 \\left\\textbar{} y \\right\\textbar-2=2 \\end{cases}$$①，或$$\\begin{cases}\\left\\textbar{} x \\right\\textbar+1=2 \\left\\textbar{} y \\right\\textbar-2=1 \\end{cases}$$②， 解①得$$\\begin{cases}x=0 y=4 \\end{cases}$$，或$$\\begin{cases}x=0 y=-4 \\end{cases}$$， 解②得$$\\begin{cases}x=1 y=3 \\end{cases}$$，或$$\\begin{cases}x=1 y=-3 \\end{cases}$$，或$$\\begin{cases}x=-1 y=-3 \\end{cases}$$，或$$\\begin{cases}x=-1 y=3 \\end{cases}$$． 综上所述，原方程组的整数解有$$6$$组． 故选$$\\text{C}$$． " ]

[ "2018年第29届希望杯初一竞赛初赛第8题4分" ]

[ [ { "aoVal": "A", "content": "$$a+b+c$$是偶数 " } ], [ { "aoVal": "B", "content": "$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}$$是偶数 " } ], [ { "aoVal": "C", "content": "$$a+b+c$$是$$3$$的倍数 " } ], [ { "aoVal": "D", "content": "$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}$$是$$3$$的倍数 " } ], [ { "aoVal": "E", "content": "$$a+b+c$$是质数 " } ] ]

[ "竞赛->知识点->数论->整除->整除的概念与基本性质", "竞赛->知识点->数论->整除->素数与合数", "课内体系->知识点->数->有理数->有理数的概念及分类->有理数的分类", "课内体系->知识点->数->有理数->有理数基础运算->有理数加法->有理数加法运算", "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", "课内体系->思想->整体思想" ]

[ "$$\\text{A}$$选项：因为$$a$$，$$b$$，$$c$$是三个大于$$3$$的质数， 所以$$a$$，$$b$$，$$c$$一定均为奇数， 所以$$a+b+c$$是奇数．错误． $$\\text{B}$$选项：因为$$a$$，$$b$$，$$c$$均为奇数， 所以$$a^{2}$$、$$b^{2}$$，$$c^{2}$$也为奇数， 所以$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}$$是奇数．错误． $$\\text{C}$$选项：若$$a=5$$、$$b=7$$、$$c=11$$， 则$$a+b+c=23$$，不是$$3$$的倍数．错误． $$\\text{D}$$选项：因为除了$$3$$，所有质数的平方除以$$3$$余数均为$$1$$， 所以$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}$$是$$3$$的倍数，正确． 故选$$\\text{D}$$. " ]

[ "1997年第14届全国初中数学联赛竞赛" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ] ]

[ "竞赛->知识点->几何杂题->几何最值" ]

[ "观察出这些距离都是$$2$$的$$n$$次幂，而$$1+2\\textless{}4$$，$$1+2+4\\textless{}8$$，$$\\cdots $$，$$1+2+4+\\cdots +16\\textless{}32$$，即这些线段中的任几条都不能构成一个多边形，这六条线段也不能构成六边形，所以至少要$$7$$个点才能满足条件． 选择$$\\text{D}$$． " ]

[ "初一单元测试《因式分解的应用》第24题", "2005年竞赛第4题6分", "初三上学期其它" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$20$$ " } ], [ { "aoVal": "C", "content": "$$24$$ " } ], [ { "aoVal": "D", "content": "$$25$$ " } ] ]

[ "课内体系->知识点->数->有理数->有理数运算巧解->裂项", "课内体系->能力->运算能力" ]

[ "∵$$\\frac{1}{{{a}^{2}}-4}=\\frac{1}{(a+2)(a-2)}=\\frac{1}{4}\\left( \\frac{1}{a+2}-\\frac{1}{a-2} \\right)$$， ∴$$A=48\\times \\left( \\frac{1}{{{3}^{2}}-4}+\\frac{1}{{{4}^{2}}-4}+\\cdots +\\frac{1}{{{100}^{2}}-4} \\right)$$ $$=48\\times \\frac{1}{4}\\left( \\frac{1}{3-2}-\\frac{1}{3+2}+\\frac{1}{4-2}-\\frac{1}{4+2}+\\frac{1}{5-2}-\\frac{1}{5+2}+\\cdots +\\frac{1}{100-2}-\\frac{1}{100+2} \\right)$$ $$=48\\times \\frac{1}{4}\\left( 1-\\frac{1}{5}+\\frac{1}{2}-\\frac{1}{6}+\\frac{1}{3}-\\frac{1}{7}+\\cdots +\\frac{1}{98}-\\frac{1}{102} \\right)$$ $$=12\\times \\left( 1+\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}-\\frac{1}{99}-\\frac{1}{100}-\\frac{1}{101}-\\frac{1}{102} \\right)$$ $$=12\\times \\left( \\frac{25}{12}-\\frac{1}{99}-\\frac{1}{100}-\\frac{1}{101}-\\frac{1}{102} \\right)$$． ∵$$\\frac{1}{99}+\\frac{1}{100}+\\frac{1}{101}+\\frac{1}{102}\\textless{}\\frac{4}{99}\\textless{}\\frac{4}{96}=\\frac{1}{24}$$， ∴$$\\frac{25}{12}\\textgreater\\frac{25}{12}-\\frac{1}{99}-\\frac{1}{100}-\\frac{1}{101}-\\frac{1}{102}\\textgreater\\frac{25}{12}-\\frac{1}{24}=\\frac{49}{24}$$， ∴$$12\\times \\frac{25}{12}\\textgreater12\\times \\left( \\frac{25}{12}-\\frac{1}{99}-\\frac{1}{100}-\\frac{1}{101}-\\frac{1}{102} \\right)\\textgreater12\\times \\frac{49}{24}$$． 即$$25\\textgreater A\\textgreater24.5$$， ∴与$$A$$最接近的正整数是$$25$$． " ]

[ "1991年第2届希望杯初二竞赛第10题" ]

[ [ { "aoVal": "A", "content": "$$2213$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{58}{21}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{2402}{49}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{365}{38}$$ " } ] ]

[ "课内体系->能力->运算能力", "课内体系->思想->方程思想", "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" ]

[ "由$$a+b=21$$，$$a$$，$$b$$是质数可知$$a$$，$$b$$必为$$2$$与$$19$$两数， 而$$\\frac{b}{a}+\\frac{a}{b}=\\frac{{{a}^{2}}+{{b}^{2}}}{ab}=\\frac{{{2}^{2}}+{{19}^{2}}}{2\\times 19}=\\frac{365}{38}$$．故选$$\\text{D}$$． " ]

[ "2018年全美数学竞赛（AMC）竞赛" ]

[ [ { "aoVal": "A", "content": "$$48$$ " } ], [ { "aoVal": "B", "content": "$$87$$ " } ], [ { "aoVal": "C", "content": "$$91$$ " } ], [ { "aoVal": "D", "content": "$$96$$ " } ], [ { "aoVal": "E", "content": "$$120$$ " } ] ]

[ "课内体系->知识点->四边形->特殊平行四边形->矩形->矩形的周长与面积", "美国AMC8->Knowledge Point->Word Problem->Word Problems Combining with Geometry->Tiles Word Problems" ]

[ "Tyler要给自己$$12$$英尺$$\\times 16$$英尺的房间铺地板．为此他打算用边长一英尺的方砖沿墙边铺，并用边长两英尺的地砖铺剩下的地面．他需要多少块地砖？ 他将在边上周围放置$$(12\\cdot 2)+(14\\cdot 2)=52$$块瓷砖．在房间的内部，我们有$$10\\cdot 14=140$$平方英尺．每个瓷砖占$$4$$个平方英尺，所以他会在房间的内部使用$$\\frac {140}{4}=35$$块瓷砖．因此，答案是$$52+35=87$$． 故选$$\\text{B}$$． He will place $$(12\\cdot 2)+(14\\cdot 2)=52$$ tiles around the border. For the inner part of the room, we have $$10\\cdot 14=140$$ square feet. Each tile takes up $$4$$ square feet, so he will use $$\\frac{140}{4}=35$$ tiles for the inner part of the room. Thus, the answer is $$52+35=87$$. " ]

[ "2013年竞赛第28题" ]

[ [ { "aoVal": "A", "content": "$$2.75$$ " } ], [ { "aoVal": "B", "content": "$$3.25$$ " } ], [ { "aoVal": "C", "content": "$$3.75$$ " } ], [ { "aoVal": "D", "content": "$$4.5$$ " } ] ]

[ "美国AMC8->Knowledge Point->Counting, Probability and Statistics->Average Problems->Average Formulas", "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" ]

[ "The average of $$1.75$$ and $$7.25$$ is equidistant from them， The average is $$(1.75 + 7.25)\\div2=4.5$$． 在数轴上，到$$1.75$$的距离和它到$$7.25$$的距离相同． $$\\text{A}$$．$$2.75$$~ ~ ~ ~ ~ $$\\text{B}$$．$$3.25$$~ ~ ~ ~ ~ $$\\text{C}$$．$$3.75$$~ ~ ~ ~ ~ $$\\text{D}$$．$$4.5$$ $$1.75$$和$$7.25$$的平均值与它们的距离相等，平均值是$$(1.75+7.25)\\div2=4.5$$． 故选$$\\text{D}$$． " ]

[ "2013年第24届全国希望杯初一竞赛复赛第4题4分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{2a-1}{9}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{3a-2}{2}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{{{a}^{2}}-6a-10}{6}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{{{a}^{2}}-2}{3}$$ " } ] ]

[ "课内体系->知识点->数->有理数->有理数的概念及分类->有理数的分类", "竞赛->知识点->数论->整除->整除的概念与基本性质" ]

[ "当$$a=5$$时，$$\\frac{2a-1}{9}=1$$，是整数； 当$$a=0$$时，$$\\frac{3a-2}{2}=-1$$，是整数； 当$$a=4$$时，$$\\frac{{{a}^{2}}-6a-10}{6}=-5$$，是整数． 而对任意整数$$a$$，$${{a}^{2}}$$是被$$3$$整除或被$$3$$除余$$1$$的整数， 所以$${{a}^{2}}-2$$是被$$3$$除余$$1$$或余$$2$$的整数， 因此对任意整数$$a$$，$$\\frac{{{a}^{2}}-2}{3}$$都不是整数． " ]

[ "2015年第26届全国希望杯初三竞赛初赛第10题" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->式->整式的加减->整式的加减运算->整式加减的综合", "课内体系->知识点->式->整式的乘除->幂的运算->幂的综合运算", "课内体系->知识点->式->整式的乘除->整式乘除化简求值" ]

[ "∵$${{x}^{2}}y-{{y}^{2}}z+{{z}^{2}}x-{{x}^{2}}z+{{y}^{2}}x+{{z}^{2}}y-2xyz=(y+x)(z-x)(z-y)$$， 且代数式的值为质数，$$x$$，$$y$$，$$z$$都是正整数， ∴$$x+y\\geqslant 2$$， ∴$$\\left { \\begin{matrix}z-x=1 z-y=1 \\end{matrix} \\right.$$， 即$$x=y$$， 且$$x+y$$为质数， 又$$x+y=2x$$， ∴$$x=y=1$$， ∴$$z=x+1=2$$， ∴$${{z}^{x+y}}={{2}^{2}}=4$$． " ]

[ "1991年第2届希望杯初二竞赛第4题" ]

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

[ "竞赛->知识点->数与式->数的运算->实数与根式运算问题" ]

[ "因为$$a=1$$，$$b=-1$$，$$c=1$$． 故选$$\\text{C}$$． " ]

[ "1998年第9届希望杯初二竞赛第1试第5题" ]

[ [ { "aoVal": "A", "content": "$$x\\ne 0$$ " } ], [ { "aoVal": "B", "content": "$$x\\ne 1$$且$$x\\ne 0$$ " } ], [ { "aoVal": "C", "content": "$$x\\ne 0$$或$$x\\ne \\pm 1$$ " } ], [ { "aoVal": "D", "content": "$$x\\ne 0$$且$$x\\ne \\pm 1$$ " } ] ]

[ "竞赛->知识点->数与式->分式->分式的基本运算" ]

[ "要使分式$$\\frac{1}{\\frac{1-\\left\\textbar{} x \\right\\textbar}{\\left\\textbar{} x \\right\\textbar}}$$有意义， 必须使$$\\left\\textbar{} x \\right\\textbar\\ne 0$$， 且$$\\frac{1-\\left\\textbar{} x \\right\\textbar}{\\left\\textbar{} x \\right\\textbar}\\ne 0$$， 即$$x\\ne 0$$且$$1-\\left\\textbar{} x \\right\\textbar\\ne 0$$， 所以$$x$$的取值范围是$$x\\ne 0$$且$$x\\ne \\pm 1$$． 故选$$\\text{D}$$． " ]

[ "1991年第2届希望杯初二竞赛第7题" ]

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

[ "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" ]

[ "当$$\\textbar x\\textbar\\textgreater1$$时，有方程$${{x}^{2}}-1=\\frac{1}{10}\\left( x+\\frac{9}{10} \\right)$$，由根与系数关系可知，方程有一正根一负根，且正根符合要求；当$$\\textbar x\\textbar\\textless{}1$$时，有方程$$1-{{x}^{2}}=\\frac{1}{10}\\left( x+\\frac{9}{10} \\right)$$，同理可知也是一正根一负根，正根符合要求，所以共有$$2$$个根． 故选$$\\text{C}$$． " ]

[ "初二上学期单元测试《分式》分式的运算第14题", "1996年第7届希望杯初二竞赛第10题" ]

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

[ "课内体系->知识点->数->有理数->有理数比较大小", "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" ]

[ "∵$$x$$、$$y$$、$$a$$、$$b$$都是正数，且$$a\\textless{}b$$， 由$$\\frac{x}{y}=\\frac{a}{b}\\textless{}1$$，得$$x\\textless{}y$$． ∴$$x$$，$$y$$中较大的数是$$y$$． 又$$x+y=c$$，$$x=\\frac{a}{b}y$$， ∴$$\\frac{a}{b}y+y=c$$， 得$$\\frac{a+b}{c}y=c$$， ∴$$y=\\frac{bc}{a+b}$$． " ]

[ "1998年第9届希望杯初一竞赛第2题" ]

[ [ { "aoVal": "A", "content": "$$-\\frac{17}{60}$$ " } ], [ { "aoVal": "B", "content": "$$-\\frac{7}{60}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{17}{60}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{7}{60}$$ " } ] ]

[ "竞赛->知识点->数与式->数的运算->有理数运算问题" ]

[ "$$a=\\left( -\\frac{1}{6} \\right)+\\left( -\\frac{1}{5} \\right)-\\left( -\\frac{1}{4} \\right)$$ $$\\textasciitilde\\textasciitilde=\\frac{(-10)+(-12)-(-15)}{60}$$ $$\\textasciitilde\\textasciitilde=\\frac{-10-12+15}{60}=-\\frac{7}{60}$$． 所以$$a$$的相反数是$$\\frac{7}{60}$$，选$$\\text{D}$$． " ]

[ "2001年第18届全国初中数学联赛竞赛第6题", "2001年第18届全国全国初中数学联赛初一竞赛" ]

[ [ { "aoVal": "A", "content": "$$522.8$$元 " } ], [ { "aoVal": "B", "content": "$$510.4$$元 " } ], [ { "aoVal": "C", "content": "$$560.4$$元 " } ], [ { "aoVal": "D", "content": "$$472.8$$元 " } ] ]

[ "课内体系->知识点->数->有理数->有理数与实际问题", "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" ]

[ "显然，$$168$$小于$$200\\times 0.9=180$$ ，没有经过打折；$$423$$小于$$500\\times 0.9=450$$ ，且大于$$200$$，所以这是经过$$9$$折后的价格；合在一起是$$168+423\\div 0.9=638\\textgreater500$$ ，按照③，可是应付款为：$$500\\times 0.9+138\\times 0.8=560.4$$ 元． 故选$$\\text{C}$$． " ]

[ "2014年第31届全国全国初中数学联赛竞赛第2题7分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{4}{7}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{5}{9}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{9}{16}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{12}{25}$$ " } ], [ { "aoVal": "E", "content": "答案请写在答题纸上 " } ] ]

[ "竞赛->知识点->数与式->因式分解->因式分解：添项、拆项" ]

[ "答案请写在答题纸上 " ]

[ "1988年全美数学竞赛（AMC）竞赛" ]

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

[ "美国AMC8->Knowledge Point->Counting, Probability and Statistics->Average Problems->Complex Average Problems", "课内体系->知识点->统计与概率->数据的分析->平均数/加权平均数" ]

[ "数列$$ {3,6,9,10 }$$中增加第五个数$$n$$，使得五个数的平均数等于其中位数．$$n$$可能值的个数是？ 添加$$n$$后可能的中间值为$$6$$、$$n$$或$$9$$．现在我们举例分析． 例$$1$$：中位数是$$6$$， 在本例中，$$n\\textless{}6$$，$$\\frac{3+n+6+9+10}{5}=6\\Rightarrow n=2$$，因此本例贡献了$$1$$． 例$$2$$：中位数是$$n$$， 我们有$$6\\textless{}n\\textless{}9$$，$$\\frac{3+6+n+9+10}{5}=n\\Rightarrow n=7$$，所以本例也贡献了$$1$$． 例$$3$$：中位数是$$9$$， 我们有$$9\\textless{}n$$，$$\\frac{3+6+9+n+10}{5}=9\\Rightarrow 17$$，所以本例也贡献$$1$$． 总共有$$3$$个可能的$$n$$值． 故选$$\\text{C}$$． The possible medians after $$n$$ is added are $$6$$, $$n$$, or $$9$$. Now we use casework Case$$1$$: The median is $$6$$, In this case, $$n\\textless{}6$$ and $$\\frac{3+n+6+9+10}{5}=6\\Rightarrow n=2$$, so this case contributes $$1$$. Case$$2$$: The median is $$n$$, We have $$6\\textless{}n\\textless{}9$$ and $$\\frac{3+6+n+9+10}{5}=n\\Rightarrow n=7$$, so this case also contributes$$1$$. Case$$3$$: The median is $$9$$, We have $$9\\textless{}n$$ and $$\\frac{3+6+9+n+10}{5}=9\\Rightarrow 17$$, so this case adds $$1$$. " ]

[ "1997年第8届希望杯初二竞赛第2试第5题", "2020~2021学年安徽合肥包河区初一下学期期末第10题3分", "2016~2017学年河南洛阳新安县初一下学期期末第8题4分" ]

[ [ { "aoVal": "A", "content": "$$M\\textgreater N\\textgreater P$$ " } ], [ { "aoVal": "B", "content": "$$N\\textgreater P\\textgreater M$$ " } ], [ { "aoVal": "C", "content": "$$P\\textgreater M\\textgreater N$$ " } ], [ { "aoVal": "D", "content": "$$M\\textgreater P\\textgreater N$$ " } ] ]

[ "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" ]

[ "$$M+1=\\frac{b+c}{a}+1=\\frac{b+c+a}{a}=\\frac{1}{a}$$， 同理$$N+1=\\frac{1}{b}$$，$$P+1=\\frac{1}{c}$$， 因为$$a\\textgreater0\\textgreater b\\textgreater c$$， 所以$$\\frac{1}{a}\\textgreater\\frac{1}{c}\\textgreater\\frac{1}{b}$$． 所以$$M+1\\textgreater P+1\\textgreater N+1$$． 所以$$M\\textgreater P\\textgreater N$$． 故选$$\\text{D}$$． " ]

[ "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$$2001=1\\times 3\\times 23\\times 29$$， $$k+1$$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． " ]

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

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

[ "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" ]

[ "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$$7$$． " ]

[ "2001年第12届希望杯初二竞赛第2试第2题" ]

[ [ { "aoVal": "A", "content": "$$2x-1$$ " } ], [ { "aoVal": "B", "content": "$$2x+1$$ " } ], [ { "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->多项式除以多项式", "课内体系->知识点->式->整式的乘除->整式的乘除运算->整式乘除的综合" ]

[ "设$$y=a{{x}^{3}}+b{{x}^{2}}+cx+d$$，除以$$\\left( x-1 \\right)\\left( x-2 \\right)$$时所得的余式为$$mx+n$$，商式为$$q\\left( x \\right)$$， 当$$y=1$$时，$$\\left( x-1 \\right)\\cdot q\\left( x \\right)+m+n=1$$， 当$$y=2$$时，$$\\left( x-2 \\right)\\cdot q\\left( x \\right)+2m+n=3$$， 所以$$m=2,n=-1$$ 所以多项式$$a{{x}^{3}}+b{{x}^{2}}+cx+d$$除以$$\\left( x-1 \\right)\\left( x-2 \\right)$$时所得的余式为$$2x-1$$． 故选$$\\text{A}$$． " ]

[ "2017年浙江宁波鄞州区宁波兴宁中学自主招生第4题5分", "2020年湖南长沙天心区湘郡培粹实验中学初一竞赛初赛（9月）第5题" ]

[ [ { "aoVal": "A", "content": "$$1.2$$元 " } ], [ { "aoVal": "B", "content": "$$1.05$$元 " } ], [ { "aoVal": "C", "content": "$$0.95$$元 " } ], [ { "aoVal": "D", "content": "$$0.9$$元 " } ] ]

[ "课内体系->知识点->方程与不等式->其他方程->三元一次方程组->三元一次方程组的实际应用" ]

[ "设一支铅笔、一本练习本和一支圆珠笔的单价分别为$$x$$、$$y$$和$$z$$元， 根据题意得：$$\\begin{cases}3x+7y+z=3.15① 4x+10y+z=4.2 ②\\end{cases}$$， $$3\\times$$①$$-$$$$2\\times$$②可得：$$x+y+z=1.05$$． 故选：$$\\text{B}$$． " ]

[ "2001年第12届希望杯初一竞赛第2试第6题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{2b}{a}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{4b}{a}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{6b}{a}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{8b}{a}$$ " } ] ]

[ "课内体系->知识点->三角形->等腰三角形->等边三角形->等边三角形的性质", "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->三角形的周长与面积问题", "课内体系->知识点->三角形->三角形及多边形->多边形->求多边形周长与面积", "课内体系->能力->推理论证能力" ]

[ "等边三角形周长为$$a$$，则边长为$$\\frac{a}{3}$$， 设$$P$$到等边三角形的三边分别为$$x$$、$$y$$、$$z$$， 则等边三角形的面积为$$b=\\frac{1}{2}\\times \\frac{a}{3}\\times \\left( x+y+z \\right)$$ 解得$$x+y+z=\\frac{6b}{a}$$， 故选C " ]

[ "2015年第32届全国全国初中数学联赛竞赛B卷第1题7分" ]

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

[ "竞赛->知识点->数与式->分式->分式的基本运算" ]

[ "∵$$a+b+c=6$$，$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=4$$， ∴$$\\frac{{{a}^{2}}+{{b}^{2}}}{2-c}+\\frac{{{b}^{2}}+{{c}^{2}}}{2-a}+\\frac{{{c}^{2}}+{{a}^{2}}}{2-b}$$ $$=\\frac{4-{{c}^{2}}}{2-c}+\\frac{4-{{a}^{2}}}{2-a}+\\frac{4-{{b}^{2}}}{2-b}$$ $$=2+c+2+a+2+b$$ $$=a+b+c+6$$ $$=6+6$$ $$=12$$． " ]