Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "初一上学期单元测试《走进美妙的数学世界》第7题", "2013年第24届全国希望杯初一竞赛初赛第3题4分" ]

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

[ "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" ]

[ "解法一： $$a=\\frac{999}{2011}=1-\\frac{1002}{2011}$$， $$b=\\frac{1000}{2012}=1-\\frac{1002}{2012}$$， $$c=\\frac{1001}{2013}=1-\\frac{1002}{2013}$$． ∵$$2011\\textless{}2013\\textless{}2013$$， ∴$$\\frac{1002}{2011}\\textgreater\\frac{1002}{2012}\\textgreater\\frac{1002}{2013}$$， ∴$$1-\\frac{1002}{2011}\\textless{}1-\\frac{1002}{2012}\\textless{}1-\\frac{1002}{2013}$$， ∴$$a\\textless{}b\\textless{}c$$． 解法二： 观察已知三个分数的特征，可设他们分别为$$\\frac{y}{x}$$，$$\\frac{y+1}{x+1}$$，$$\\frac{y+2}{x+2}$$（$$x\\textgreater y\\textgreater0$$）， 则$$\\frac{y}{x}-\\frac{y+1}{x+1}=\\frac{y(x+1)-x(y+1)}{x(x+1)}=\\frac{y-x}{x(x+1)}$$， ∵$$x\\textgreater y\\textgreater0$$， ∴$$y-x\\textless{}0$$， ∴$$\\frac{y-x}{x(x+1)}\\textless{}0$$， ∴$$\\frac{y}{x}\\textless{}\\frac{y+1}{x+1}$$． 同理$$\\frac{y+1}{x+1}\\textless{}\\frac{y+2}{x+2}$$， ∴$$\\frac{y}{x}\\textless{}\\frac{y+1}{x+1}\\textless{}\\frac{y+2}{x+2}$$， ∴$$a\\textless{}b\\textless{}c$$． " ]

[ "1996年第7届希望杯初二竞赛第8题4分" ]

[ [ { "aoVal": "A", "content": "$$12$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "课内体系->知识点->式->分式->分式的运算->分式加减混合运算", "课内体系->知识点->式->分式->分式的运算->分式通分" ]

[ "原式$$=\\frac{2x-6-2x-6+2x+18}{(x+3)(x-3)}=\\frac{2(x+3)}{(x+3)(x-3)}=\\frac{2}{x-3}$$ $$\\therefore x-3=1$$，$$-1$$，$$-2$$，$$2$$时，原式的值为整数 此时$${{x}_{1}}=4$$，$${{x}_{2}}=2$$，$${{x}_{3}}=1$$，$${{x}_{4}}=5$$ $$\\therefore {{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=12$$． " ]

[ "1993年第4届全国希望杯初一竞赛复赛第7题" ]

[ [ { "aoVal": "A", "content": "$$148$$ " } ], [ { "aoVal": "B", "content": "$$247$$ " } ], [ { "aoVal": "C", "content": "$$93$$ " } ], [ { "aoVal": "D", "content": "$$122$$ " } ] ]

[ "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->有理数乘方->含乘方的有理数混合运算", "课内体系->知识点->数->有理数->有理数基础运算->有理数除法->有理数四则混合运算" ]

[ "$$n\\left( n+1 \\right)$$为偶数． 设$$302$$被$$n\\left( n+1 \\right)$$除商$$q$$余$$r$$，则$$302=n\\left( n+1 \\right)q+r$$知，$$r$$为偶数．显然$$\\text{B}$$、$$\\text{C}$$均应排除． 由除数$$n\\left( n+1 \\right)$$只能取$$6$$，$$12$$，$$20$$，$$30$$，$$42$$，$$56$$，$$72$$，$$90$$，$$110$$，$$132$$，$$156$$，$$182$$，$$210$$，$$240$$，$$272$$这些值，计算得相应的余数中最小的正值为$$2$$，最大正值为$$146$$， 所以$$r$$的正的最小值与最大值的和是$$148$$．选$$\\text{A}$$． " ]

[ "2010年第21届全国希望杯初一竞赛初赛第1题4分" ]

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

[ "知识标签->知识点->数->有理数->正数和负数->正数和负数的定义", "知识标签->题型->数->有理数->数轴与有理数有关的概念->题型：区分正负数" ]

[ "因为$$a\\textless{}0$$， 所以$$\\left\\textbar{} a \\right\\textbar$$，$$-a$$，$${{a}^{2010}}$$，$$\\left\\textbar{} -a \\right\\textbar$$均为正数， 而$$\\left( \\frac{{{a}^{2}}}{a}-a \\right)=a-a=0$$，$$\\left( \\frac{{{a}^{2}}}{a}+a \\right)=a+a=2a$$， 所以只有$${{a}^{2009}}$$，$$\\left( \\frac{{{a}^{2}}}{a}+a \\right)$$为负数． 故负数的个数是$$2$$． " ]

[ "2014年第25届全国希望杯初一竞赛初赛第2题4分" ]

[ [ { "aoVal": "A", "content": "平行四边形 " } ], [ { "aoVal": "B", "content": "圆 " } ], [ { "aoVal": "C", "content": "等边三角形 " } ], [ { "aoVal": "D", "content": "正方形 " } ] ]

[ "课内体系->知识点->几何变换->轴对称->轴对称基础->对称轴条数问题" ]

[ "平行四边形不一定是轴对称图形； 圆有无数条对称轴； 等边三角形恰好有三条对称轴； 正方形有四条对称轴． 故选$$\\text{C}$$． " ]

[ "1993年第4届全国希望杯初一竞赛初赛第13题" ]

[ [ { "aoVal": "A", "content": "$$-\\frac{171}{40}$$ " } ], [ { "aoVal": "B", "content": "$$-\\frac{347}{80}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{71}{220}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{142}{55}$$ " } ] ]

[ "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->常规一元一次方程解法", "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->能力->运算能力" ]

[ "以$$\\frac{1}{2}$$，$$\\frac{11}{5}$$，$$8$$分别代入所给三个方程知： $$x=\\frac{1}{2}$$是方程$$\\frac{2x-1}{3}-\\frac{10x+1}{12}=\\frac{2x+1}{4}-1$$的根， $$y=8$$是方程$$3(2y+1)=2(1+y)+3(y+3)$$的根， $$z=\\frac{11}{5}$$是方程$$\\frac{1}{2}\\left[ z-\\frac{1}{2}(z-1) \\right]=\\frac{2}{3}(z-1)$$的根， ∴$$\\frac{x}{y}-\\frac{z}{x}=\\frac{\\frac{1}{2}}{8}-\\frac{\\frac{11}{5}}{\\frac{1}{2}}=\\frac{1}{16}-\\frac{22}{5}=-\\frac{347}{80}$$，选$$\\text{B}$$． " ]

[ "2014年第25届全国希望杯初二竞赛复赛第2题4分", "2015~2016学年湖南长沙雨花区湖南广益实验中学初一上学期期中" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$-3$$ " } ], [ { "aoVal": "C", "content": "$$-4$$ " } ], [ { "aoVal": "D", "content": "$$-5$$ " } ] ]

[ "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" ]

[ "当$$x=1$$时，代数式$$a{{x}^{3}}+bx+1$$的值是$$5$$， 所以$$a+b+1=5$$，即$$a+b=4$$． 当当$$x=-1$$时，$$a{{x}^{3}}+bx+1=-(a+b)+1=-4+1=-3$$． " ]

[ "1991年第2届全国希望杯初一竞赛初赛第12题" ]

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

[ "课内体系->知识点->式->分式->分式的基础->分式为特殊值", "课内体系->知识点->式->分式->分式化简求值->分式化简求值-直接代入数值求值" ]

[ "$$\\frac{4x-2y}{xy}=\\frac{4\\times \\frac{1}{2}-2\\times (-2)}{\\frac{1}{2}\\times (-2)}=\\frac{2+4}{-1}=-6$$．选$$\\text{A}$$． " ]

[ "2013年竞赛第5题4分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{607}{967}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{1821}{967}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{5463}{967}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{16389}{967}$$ " } ] ]

[ "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" ]

[ "设$$2013*2012*\\cdots *4=m$$， 则$$\\left( 2013*2012*\\cdots *4 \\right)*3=m*3$$， $$=\\frac{3{{m}^{3}}\\times 3+3{{m}^{2}}\\times 9+m\\times 27+45}{{{m}^{3}}+3{{m}^{2}}+3m+1+64-60}=9$$， 于是$$\\left( 2013*2012*\\cdots *3 \\right)*2=9*2$$， $$=\\frac{3\\times {{9}^{3}}\\times 2+3\\times {{9}^{2}}\\times {{2}^{2}}+9\\times {{2}^{3}}+45}{{{10}^{3}}+{{3}^{3}}-60}$$， $$=\\frac{5463}{967}$$． 故选$$\\text{C}$$． " ]

[ "2019年浙江杭州拱墅区杭州市文澜中学初二竞赛第10题3分" ]

[ [ { "aoVal": "A", "content": "$$x ~\\textless{} ~-1$$ " } ], [ { "aoVal": "B", "content": "$$x\\textgreater-1$$ " } ], [ { "aoVal": "C", "content": "$$x ~\\textless{} ~1$$ " } ], [ { "aoVal": "D", "content": "$$x\\textgreater1$$ " } ] ]

[ "课内体系->知识点->函数->一次函数->一次函数与方程、不等式->一次函数与一元一次不等式" ]

[ "∵$$y=kx+2(k\\textgreater0)$$中， $$k\\textgreater0$$． ∴$$y$$随$$x$$增大而增大， ∵$$({{x}_{1}},{{y}_{1}})$$$$({{x}_{2}},{{y}_{2}})$$是$$y=kx+2$$图象上不同两点， ∴$${{x}_{1}}\\textgreater{{x}_{2}}$$时$${{y}_{1}}\\textgreater{{y}_{2}}$$， $${{x}_{1}} ~\\textless{} ~{{x}_{2}}$$时，$${{y}_{1}} ~\\textless{} ~{{y}_{2}}$$． ∴$$({{x}_{2}}-{{x}_{1}})({{y}_{1}}-{{y}_{2}}) ~\\textless{} ~0$$， ∴$$m=({{x}_{2}}-{{x}_{1}})({{y}_{1}}-{{y}_{2}}) ~\\textless{} ~0$$． ∴由$$mx ~\\textless{} ~-m$$， 解得$$x\\textgreater-1$$． 故选$$\\text{B}$$． " ]

[ "2019~2020学年3月浙江杭州滨江区杭州二中白马湖学校初一下学期周测C卷第9题", "第7届希望杯初二竞赛第3题", "2017~2018学年5月四川成都天府新区 成都市实验外国语学校（西区）初二下学期月考第10题3分", "上海自主招生近5年（2015-2019）真题题型分类汇编第15题", "初二其它" ]

[ [ { "aoVal": "A", "content": "$$3$$个 " } ], [ { "aoVal": "B", "content": "$$4$$个 " } ], [ { "aoVal": "C", "content": "$$6$$个 " } ], [ { "aoVal": "D", "content": "$$8$$个 " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->式->因式分解->因式分解的基础->已知因式分解结果求参数" ]

[ "设$$x^{2}+ax-12$$能分解成两个整数系数的一次因式的乘积， 即$$x^{2}+ax-12=(x+m)(x+n)$$，$$m$$，$$n$$是整数， $$\\therefore x^{2}+ax-12=x^{2}+(m+n)x+mn$$， $$\\therefore \\begin{cases}mn=-12 \\m+n=a\\end{cases}$$， $$\\because m$$，$$n$$是整数，且$$mn=-12$$， 有$$\\begin{cases}m=12 \\n=-1\\end{cases}$$，$$\\begin{cases}m=-1 \\n=12\\end{cases}$$，$$\\begin{cases}m=6 \\n=-2\\end{cases}$$，$$\\begin{cases}m=-2 \\n=6\\end{cases}$$， $$\\begin{cases}m=4 \\n=-3\\end{cases}$$，$$\\begin{cases}m=-3 \\n=4\\end{cases}$$，$$\\begin{cases}m=3 \\n=-4\\end{cases}$$，$$\\begin{cases}m=-3 \\n=4\\end{cases}$$，$$\\begin{cases}m=2 \\n=-6\\end{cases}$$， $$\\begin{cases}m=-6 \\n=2\\end{cases}$$，$$\\begin{cases}m=1 \\n=-12\\end{cases}$$，$$\\begin{cases}m=-12 \\n=1\\end{cases}$$，共$$12$$种情况． 而$$a=m+n$$，只有$$6$$种结果， 故选$$\\text{C}$$． " ]

[ "1995年第6届希望杯初二竞赛第1题" ]

[ [ { "aoVal": "A", "content": "$${{x}_{0}}\\textgreater2{{x}_{0}}\\textgreater x_{0}^{2}$$ " } ], [ { "aoVal": "B", "content": "$$x_{0}^{2}\\textgreater{{x}_{0}}\\textgreater2{{x}_{0}}$$ " } ], [ { "aoVal": "C", "content": "$$x_{0}^{2}\\textgreater2{{x}_{0}}\\textgreater{{x}_{0}}$$ " } ], [ { "aoVal": "D", "content": "$$2{{x}_{0}}\\textgreater x_{0}^{2}\\textgreater{{x}_{0}}$$ " } ] ]

[ "竞赛->知识点->方程与不等式->一次方程->一元一次方程" ]

[ "原方程变为$$\\left\\textbar{} \\frac{1+x}{2} \\right\\textbar=\\left\\textbar{} x \\right\\textbar$$，即$$\\frac{1+x}{2}=\\pm x$$，解得$$x=1$$或$$x=-\\frac{1}{3}$$，由题意知$${{x}_{0}}=-\\frac{1}{3}$$，于是$$x_{0}^{2}\\textgreater{{x}_{0}}\\textgreater2{{x}_{0}}$$，故选$$\\text{B}$$． " ]

[ "2007年第18届希望杯初一竞赛复赛第9题4分" ]

[ [ { "aoVal": "A", "content": "$$2$$厘米 " } ], [ { "aoVal": "B", "content": "$$3$$厘米 " } ], [ { "aoVal": "C", "content": "$$6$$厘米 " } ], [ { "aoVal": "D", "content": "$$7$$厘米 " } ] ]

[ "课内体系->能力->运算能力", "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->认识图形" ]

[ "由构成的长方体的底面矩形周长为$$18$$，得长$$+$$宽$$=9$$厘米， 因为长方体的体积为$$42$$立方厘米， 所以长和宽都应是$$42$$的约数，且底面面积也是$$42$$的约数， 又$$42=1\\times 42=2\\times 3\\times 7$$， 所以长$$=7$$厘米，宽$$=2$$厘米， 此时高$$=42\\div (2\\times 7)=3$$（厘米）. 故选($$\\text{B}$$)． " ]

[ "1992年第3届全国希望杯初一竞赛复赛第8题" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$${{1992}^{2}}$$ " } ], [ { "aoVal": "C", "content": "$${{2}^{1992}}$$ " } ], [ { "aoVal": "D", "content": "$${{4}^{1992}}$$ " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->式->整式的加减->整式有关的概念->整式->整式的定义", "课内体系->知识点->式->整式的乘除->幂的运算->幂的综合运算" ]

[ "由$$x-y=2$$,平方得$${{x}^{2}}-2xy+{{y}^{2}}=4$$． 又已知$${{x}^{2}}+{{y}^{2}}=4$$， 两式相减得$$2xy=0$$，∴$$xy=0$$． 所以$$x$$，$$y$$中至少有一个为$$0$$，但$${{x}^{2}}+{{y}^{2}}=4$$． 因此，$$x$$，$$y$$中只能有一个为$$0$$，另一个为$$2$$或$$-2$$． 无论哪种情况，都有$${{x}^{1992}}+{{y}^{1992}}={{0}^{1992}}+{{(\\pm 2)}^{1992}}={{2}^{1992}}$$，选$$\\text{C}$$． " ]

[ "2010年第21届希望杯初二竞赛第1试第10题" ]

[ [ { "aoVal": "A", "content": "$$32$$秒 " } ], [ { "aoVal": "B", "content": "$$38$$秒 " } ], [ { "aoVal": "C", "content": "$$42$$秒 " } ], [ { "aoVal": "D", "content": "$$48$$秒 " } ] ]

[ "竞赛->知识点->方程与不等式->一次方程->一元一次方程" ]

[ "设从楼上到楼下的路程为$$s$$，则自动扶梯每秒向下运动$$\\frac{s}{56}$$的路程． 人沿向下运动的扶梯向下走，经过$$24$$秒，扶梯向下运动了$$24\\times \\frac{s}{56}=\\frac{3s}{7}$$的路程， 人向下走了$$s-\\frac{3}{7}s=\\frac{4}{7}s$$的路程， 所以，人每秒向下走$$\\frac{4}{7}s\\div 24=\\frac{s}{42}$$的路程， 那么，人走$$s$$的路程要用$$42$$秒， 即人沿不动的扶梯从楼上走到楼下要用$$42$$秒． 故选$$\\text{C}$$． " ]

[ "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9题3分" ]

[ [ { "aoVal": "A", "content": "$$3.19\\times {{10}^{6}}$$ " } ], [ { "aoVal": "B", "content": "$$3.19\\times {{10}^{8}}$$ " } ], [ { "aoVal": "C", "content": "$$0.319\\times {{10}^{7}}$$ " } ], [ { "aoVal": "D", "content": "$$319\\times {{10}^{6}}$$ " } ] ]

[ "课内体系->知识点->数->有理数->科学记数法->科学记数法：表示较大的数" ]

[ "$$319$$万$$=3190000$$， $$3190000=3.19\\times {{10}^{6}}$$， 其中科学记数法的表示方法为$$\\left\\textbar{} a \\right\\textbar\\times {{10}^{n}}\\left( 1\\leqslant a\\textless{}10 \\right)$$． 故选$$\\text{A}$$． " ]

[ "2009年第20届希望杯初一竞赛第1试第8题4分" ]

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

[ "竞赛->知识点->三角形->三角形基础" ]

[ "由面积法知：点$$P$$到等边三角形三边的距离之和等于这个等边三角形的高，而$$b=\\frac{1}{2}\\cdot \\frac{a}{3}\\cdot h$$，所以$$h=\\frac{6b}{a}$$． 故选$$\\text{C}$$． " ]

[ "1998年第9届希望杯初二竞赛第2试第7题" ]

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

[ "课内体系->思想->方程思想", "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->加减消元法解二元一次方程组", "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->解可化为一元一次方程的分式方程" ]

[ "因为$$\\frac{m}{x+3}-\\frac{n}{x-3}=\\frac{8x}{{{x}^{2}}-9}$$， 所以$$\\frac{m(x-3)-n(x+3)}{{{x}^{2}}-9}=\\frac{8x}{{{x}^{2}}-9}$$． 因为$$x\\ne \\pm 3$$． 所以$$m(x-3)-n(x+3)=8x$$． 即$$(m-n)x-3(m+n)=8x$$． 此式对于一切$$x\\ne \\pm 3$$的值均成立． 所以$$\\begin{cases}m-n=8, m+n=0, \\end{cases}$$ 解得$$\\begin{cases}m=4, n=-4. \\end{cases}$$ 所以$$mn=-16$$． 故选$$\\text{D}$$． " ]

[ "2019~2020学年10月山西太原杏花岭区太原志达中学初三上学期月考第2题3分", "2019~2020学年10月广东佛山禅城区佛山市惠景中学初三上学期月考第6题3分", "2019~2020学年9月甘肃兰州城关区兰州外国语学校初三上学期月考第8题4分", "2020~2021学年9月四川成都天府新区天府七中初三上学期周测B卷第8题3分", "2020~2021学年10月广东深圳南山区深圳南山外国语学校高新中学初三上学期周测B卷第4题", "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2题4分", "2018~2019学年辽宁沈阳浑南区育才实验学校初二下学期期末第3题", "2020~2021学年9月四川成都天府新区天府七中初三上学期周测C卷第8题3分" ]

[ [ { "aoVal": "A", "content": "$$k\\textgreater\\frac{1}{4}$$且$$k\\ne 0$$ " } ], [ { "aoVal": "B", "content": "$$k\\textless{}\\frac{1}{4}$$且$$k\\ne 0$$ " } ], [ { "aoVal": "C", "content": "$$k\\leqslant \\frac{1}{4}$$且$$k\\ne 0$$ " } ], [ { "aoVal": "D", "content": "$$k\\textless{}\\frac{1}{4}$$ " } ] ]

[ "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->根的判别式的应用->由一元二次方程根的情况确定参数" ]

[ "∵关于$$x$$的一元二次方程$$k{{x}^{2}}-x+1=0$$有实数根， ∴$$k\\ne 0$$且$$\\Delta ={{b}^{2}}-4ac\\geqslant 0$$， ∴$$1-4k\\geqslant 0$$， ∴解得：$$k\\leqslant \\frac{1}{4}$$． 综上，$$k\\leqslant \\frac{1}{4}$$且$$k\\ne 0$$． 故选$$\\text{C}$$． " ]

[ "2018年浙江宁波余姚市余姚市实验学校初二竞赛第5题5分" ]

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

[ "课内体系->知识点->统计与概率->数据的分析->方差", "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" ]

[ "∵这五个数的平方和为$$200$$，平均数为$$6$$， ∴$$x_1+x_2+x_3+x_4+x_5=6\\times5=30$$， $$x_1^{2}+x_2^{2}+x_3^{3}+x_4^{4}+x_5^{2}=200$$， ∴这组数据的方差为： $$\\frac15\\times\\left[(x_1-6)^{2}+(x_2-6)^{2}+(x_3-6)^{2}+(x_4-6)^{2}+(x_5-6)^{2}\\right]$$ $$=\\frac15\\times\\left[x_1^{2}+x_2^{2}+x_3^{3}+x_4^{4}+x_5^{2}+5\\times6^{2}-12(x_1+x_2+x_3+x_4+x_5)\\right]$$ $$=\\frac15\\times(200+5\\times36-12\\times30)$$ $$=\\frac15\\times(200+180-360)$$ $$=\\frac15\\times20$$ $$=4$$， 故选：$$\\text{C}$$． " ]

[ "2015年第26届全国希望杯初一竞赛复赛第9题" ]

[ [ { "aoVal": "A", "content": "小尼 " } ], [ { "aoVal": "B", "content": "小浩 " } ], [ { "aoVal": "C", "content": "小莲 " } ], [ { "aoVal": "D", "content": "小雪 " } ] ]

[ "知识标签->知识点->命题与证明" ]

[ "用假设法： （$$1$$）假设小尼说了假话，那么苹果在小尼那里，小浩和小莲就说的是假话，与题设冲突； （$$2$$）假设小浩说的是假话，其他人说的都是真话，那么苹果在小浩那里； （$$3$$）假设小莲说了假话，则小浩说``苹果在小雪那里''与小雪说``苹果不在我这里''都是真话，互相矛盾； （$$4$$）假设小雪说的是假话，那么小莲说的就是假话，这样小雪、小莲都说假话，与``其中只有一人说的是假话''的题设不符． 因此苹果在小浩那里． " ]

[ "2015年全美数学竞赛（AMC）竞赛" ]

[ [ { "aoVal": "A", "content": "$$3024$$ " } ], [ { "aoVal": "B", "content": "$$4536$$ " } ], [ { "aoVal": "C", "content": "$$5040$$ " } ], [ { "aoVal": "D", "content": "$$6480$$ " } ], [ { "aoVal": "E", "content": "$$6561$$ " } ] ]

[ "课内体系->知识点->数->有理数->数轴->数轴上整点覆盖问题", "美国AMC8->Knowledge Point->Combinatorics->Permutation and Combination->Permutation Problems" ]

[ "$$1000$$到$$9999$$之间有多少个整数四个数位上各不相同？ 这个问题可以改为``\\,''有多少四位正整数有四个不同的数，因为数字是$$1000$$到$$9999$$之间的四位整数，对于第一个数有$$9$$中选择，因为它不能是$$0$$，第二个数只有$$9$$种选择，因为它不同于第一个数，第三个数有$$8$$种选择，因为它不同于前两个数，第四个数有$$7$$种选择，因为它不同于前三个数，则$$9\\times 9\\times 8\\times 7=4536$$． 故选$$\\text{B}$$． The question can be rephrased to \\textquotesingle How many four-digit positive integers have four distinct digits?\\textquotesingle, since numbers between $$1000$$ and $$9999$$ are four-digit integers. There are $$9$$ choices for the first number, since it cannot be $$0$$, there are only $$9$$ choices left for the second number since it must differ from the first, $$8$$ choices for the third number, since it must differ from the first two, and $$7$$ choices for the fourth number, since it must differ from all three. This means there are $$9\\times9\\times8\\times7=\\boxed{(\\text{B})4536}$$ integers between $$1000$$ and $$9999$$ with four distinct digits. " ]

[ "2010年第21届希望杯初二竞赛第1试第3题" ]

[ [ { "aoVal": "A", "content": "$$x\\leqslant 2010$$ " } ], [ { "aoVal": "B", "content": "$$x\\leqslant 2010$$，且$$x\\ne \\pm 2009$$ " } ], [ { "aoVal": "C", "content": "$$x\\leqslant 2010$$，且$$x\\ne 2009$$ " } ], [ { "aoVal": "D", "content": "$$x\\leqslant 2010$$，且$$x\\ne -2009$$ " } ] ]

[ "竞赛->知识点->数与式->分式->分式的基本运算" ]

[ "由题意，得$$\\begin{cases}2010-x\\geqslant 0 \\textbar x\\textbar-2009\\ne 0 \\end{cases}$$， 解得$$x\\leqslant 2010$$，且$$x\\ne \\pm 2009$$． 故选$$\\text{B}$$． " ]

[ "1999年第10届希望杯初一竞赛第5题" ]

[ [ { "aoVal": "A", "content": "$$9$$ " } ], [ { "aoVal": "B", "content": "$$7.5$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6.5$$ " } ] ]

[ "课内体系->思想->方程思想", "课内体系->知识点->数->有理数->相反数->相反数的性质", "课内体系->知识点->数->有理数->倒数与负倒数" ]

[ "$$7-a$$的倒数是$$\\frac{1}{7-a}$$，$$\\frac{1}{7-a}$$的相反数是$$-\\frac{1}{7-a}=\\frac{1}{a-7}$$． 依题意列方程$$\\frac{1}{a-7}=-2$$． 解得$$a=\\frac{13}{2}=6.5$$． 故选$$\\text{D}$$． " ]

[ "2004年第15届希望杯初二竞赛第1试第4题" ]

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

[ "竞赛->知识点->数与式->数的运算->有理数运算问题" ]

[ "设$$a=3$$，$$b=5$$， 则$$\\frac{a+b}{ab}=\\frac{3+5}{3\\times 5}=\\frac{8}{15}$$是最简分数． $$\\frac{b-a}{b+a}=\\frac{5-3}{5+3}=\\frac{2}{8}$$不是最简分数． $$\\frac{{{b}^{2}}-{{a}^{2}}}{{{a}^{2}}+{{b}^{2}}}==\\frac{{{5}^{2}}-{{3}^{2}}}{{{3}^{2}}+{{5}^{2}}}=\\frac{16}{34}$$不是最简分数． $$\\frac{ab}{{{a}^{2}}+{{b}^{2}}}=\\frac{3\\times 5}{{{3}^{2}}+{{5}^{2}}}=\\frac{15}{34}$$是最简分数． 故选$$\\text{B}$$． " ]

[ "2011年全美数学竞赛（AMC）竞赛" ]

[ [ { "aoVal": "A", "content": "中位数$$\\textless$$平均值$$\\textless$$众数 " } ], [ { "aoVal": "B", "content": "平均值$$\\textless$$众数$$\\textless$$中位数 " } ], [ { "aoVal": "C", "content": "平均值$$\\textless$$中位数$$\\textless$$众数 " } ], [ { "aoVal": "D", "content": "中位数$$\\textless$$众数$$\\textless$$平均值 " } ], [ { "aoVal": "E", "content": "众数$$\\textless$$中位数$$\\textless$$平均值 " } ] ]

[ "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" ]

[ "如下是 Tyler 去年夏天外出捕鱼九次的数量：$$2$$，$$0$$，$$1$$，$$3$$，$$0$$，$$3$$，$$3$$，$$1$$，$$2$$．下列关于均值，中位数，众数的描述哪个正确？ A~ 中位数$$\\textless$$均值$$\\textless$$众数 B~ 均值$$\\textless$$众数$$\\textless$$中位数 C~ 均值数$$\\textless$$中位数$$\\textless$$众数 D~ 中位数$$\\textless$$众数$$\\textless$$均值 E~ 众数$$\\textless$$中位数$$\\textless$$均值 首先，把数字按增加的排列． $$0$$，$$0$$，$$1$$，$$1$$，$$2$$，$$2$$，$$3$$，$$3$$，$$3$$ 故$$\\frac{0+0+1+1+2+2+3+3+3}{9}=\\frac{15}{9}$$，中位数是$$2$$，众数是$$3$$．因为，$$\\frac{15}{9} ~\\textless{} ~2 ~\\textless{} ~3$$，所以答案是mean$$ ~\\textless{} ~$$median$$ ~\\textless{} ~$$mode，即平均数$$ ~\\textless{} ~$$中位数$$ ~\\textless{} ~$$众数． 故选$$\\text{C}$$． First, put the numbers in increasing order. $$0$$, $$0$$, $$1$$, $$1$$, $$2$$, $$2$$, $$3$$, $$3$$, $$3$$ The mean is $$\\frac{0+0+1+1+2+2+3+3+3}{9}= \\frac{15}{9}$$, the median is $$2$$, and the mode is $$3$$. Because, $$\\frac{15}{9}\\textless2\\textless3$$, the answer is mean $$\\textless{}$$ median $$\\textless{}$$ mode. " ]

[ "2018年全国初中数学联赛竞赛B卷" ]

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

[ "知识标签->题型->方程与不等式->一元二次方程", "知识标签->题型->式->整式的乘除->幂的运算", "知识标签->学习能力->运算能力", "知识标签->学习能力->抽象概括能力", "知识标签->知识点->式->整式的乘除->幂的运算", "知识标签->知识点->方程与不等式->一元二次方程" ]

[ "当$$x+3=0$$且$$x-1\\ne 0$$时，$$x=-3$$， 当$$x-1=1$$时，$$x=2$$， 当$$x-1=-1$$且$$x+3$$为偶数时，$$x=0$$不符合题意， 所以，满足条件的整数$$x$$有$$2$$个． 故选$$\\text{B}$$． " ]

[ "1994年第5届全国希望杯初一竞赛复赛第2题" ]

[ [ { "aoVal": "A", "content": "$$\\left\\textbar{} b \\right\\textbar-\\left\\textbar{} a \\right\\textbar$$ " } ], [ { "aoVal": "B", "content": "$$-\\left\\textbar{} a \\right\\textbar-\\left\\textbar{} b \\right\\textbar$$ " } ], [ { "aoVal": "C", "content": "$$\\left\\textbar{} a \\right\\textbar-\\left\\textbar{} b \\right\\textbar$$ " } ], [ { "aoVal": "D", "content": "$$\\left\\textbar{} a \\right\\textbar+\\left\\textbar{} b \\right\\textbar$$ " } ] ]

[ "课内体系->知识点->数->有理数->绝对值->绝对值的非负性" ]

[ "因为$$a\\textless{}0$$，所以$$\\left\\textbar{} a \\right\\textbar=-a$$． 因为$$b\\textgreater0$$，所以$$\\left\\textbar{} b \\right\\textbar=b$$． 所以$$a+b=-\\left\\textbar{} a \\right\\textbar+\\left\\textbar{} b \\right\\textbar$$．选$$\\text{A}$$． " ]

[ "2019~2020学年11月浙江杭州拱墅区杭州树兰中学初一上学期周测C卷第4题3分", "2019~2020学年4月浙江温州乐清市乐清市乐成公立寄宿学校初一下学期月考（竞赛班）第7题4分", "2016~2017学年12月江苏镇江句容市句容市华阳学校初一上学期月考第17题3分", "2017~2018学年陕西西安雁塔区西安高新第一中学初中校区东区初级中学初一上学期期末第3题5分" ]

[ [ { "aoVal": "A", "content": "$$6$$折 " } ], [ { "aoVal": "B", "content": "$$7$$折 " } ], [ { "aoVal": "C", "content": "$$8$$折 " } ], [ { "aoVal": "D", "content": "$$9$$折 " } ] ]

[ "课内体系->知识点->方程与不等式->不等式（组）->不等式组应用->不等式组的经济问题", "课内体系->能力->运算能力" ]

[ "要保持利润率不低于$$5 \\%$$，设可打$$x$$折． 则$$1575\\times \\frac{x}{10}-1200\\geqslant 1200\\times 5 \\%$$， 解得$$x\\geqslant 8$$， 即要保持利润率不低于$$5 \\%$$，最多可打$$8$$折． 故选：$$\\text{C}$$． " ]

[ "2014年第25届全国希望杯初二竞赛复赛第4题4分" ]

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

[ "课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式组->一元一次不等式组的整数解", "课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式组->解一元一次不等式组", "课内体系->能力->运算能力" ]

[ "解$$x-4\\leqslant 8-2x$$，得$$x\\leqslant 4$$， 故原不等式的解为$$-\\frac{2}{3}\\textless{}x\\leqslant 4$$， 所以，原不等式组的最小整数解是$$x=0$$． " ]

[ "2010年第15届华杯赛初一竞赛初赛第1题" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{15}{2}$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

[ "竞赛->知识点->方程与不等式->一次方程->方程组" ]

[ "$$2x+3y=15$$，$$6x+13y=41$$， 由此可得：$$8x+16y=56$$，两边除以$$8$$得到$$x+2y=7$$． 故选$$\\text{B}$$． " ]

[ "1991年第2届希望杯初二竞赛第6题" ]

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

[ "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->判断能否构成三角形", "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" ]

[ "因为$$x=y\\textgreater5$$的任何正整数，都可以和$$10$$作为三角形的三条边． 故选$$\\text{D}$$． " ]

[ "1998年第9届希望杯初二竞赛第2试第4题" ]

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

[ "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" ]

[ "原式$$=\\sqrt{{{(\\sqrt{\\frac{9}{2}}+\\sqrt{\\frac{7}{2}})}^{2}}}+\\sqrt{{{(\\sqrt{\\frac{9}{2}}-\\sqrt{\\frac{7}{2}})}^{2}}}$$$$=2\\sqrt{\\frac{9}{2}}=3\\sqrt{2}$$．故选$$\\text{A}$$． " ]

[ "初一下学期其它", "1997年第8届希望杯初二竞赛第2试第6题" ]

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

[ "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" ]

[ "因为$${{m}^{2}}+m-1=0$$， 所以$${{m}^{2}}+m=1$$， 所以$${{m}^{3}}+2{{m}^{2}}-1997$$$$=m({{m}^{2}}+m)+{{m}^{2}}-1997$$$$=m+{{m}^{2}}-1997$$$$=1-1997$$$$=-1996$$． " ]

[ "2017年全国全国初中数学联赛竞赛" ]

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

[ "课内体系->知识点->函数->二次函数->二次函数与方程、不等式->二次函数与坐标轴交点", "课内体系->能力->运算能力" ]

[ "因为二次函数$$y=a{{x}^{2}}+bx+c$$的图象与$$x$$轴有唯一交点， 所以$${{\\Delta }_{1}}={{b}^{1}}-4ac=0$$， 所以$${{b}^{2}}=4ac\\ne 0$$． 故二次函数$$y={{a}^{3}}{{x}^{2}}+{{b}^{3}}x+{{c}^{3}}$$的判别式$${{\\Delta }_{2}}={{({{b}^{3}})}^{2}}-4{{a}^{3}}{{c}^{3}}={{b}^{6}}-\\frac{1}{16}{{(4ac)}^{3}}={{b}^{6}}-\\frac{1}{16}{{({{b}^{2}})}^{3}}=\\frac{15}{16}{{b}^{6}}\\textgreater0$$， 所以，二次函数$$y={{a}^{3}}{{x}^{2}}+{{b}^{3}}x+{{c}^{3}}$$的图象与$$x$$轴有两个交点． 故选$$\\text{C}$$． " ]

[ "1990年第1届希望杯初二竞赛第1题" ]

[ [ { "aoVal": "A", "content": "$$45{}^{}\\circ $$ " } ], [ { "aoVal": "B", "content": "$$75{}^{}\\circ $$ " } ], [ { "aoVal": "C", "content": "$$55{}^{}\\circ $$ " } ], [ { "aoVal": "D", "content": "$$65{}^{}\\circ $$ " } ] ]

[ "竞赛->知识点->几何图形初步->角->角的基础问题" ]

[ "因为所求角$$a=5(90{}^{}\\circ -a)$$，解得$$a=75{}^{}\\circ $$． 故选$$\\text{B}$$． " ]

[ "2013年竞赛第2题3分", "2016年安徽芜湖镜湖区芜湖市第一中学初三自主招生第5题6分" ]

[ [ { "aoVal": "A", "content": "$$-6 ~\\textless{} ~t ~\\textless{} ~-\\frac{11}{2}$$ " } ], [ { "aoVal": "B", "content": "$$-6\\leqslant t ~\\textless{} ~-\\frac{11}{2}$$ " } ], [ { "aoVal": "C", "content": "$$-6 ~\\textless{} ~t\\leqslant -\\frac{11}{2}$$ " } ], [ { "aoVal": "D", "content": "$$-6\\leqslant t\\leqslant -\\frac{11}{2}$$ " } ] ]

[ "竞赛->知识点->方程与不等式->不等式->一次不等式组" ]

[ "根据题设知不等式组有解，解得，$$3-2t ~\\textless{} ~x ~\\textless{} ~20$$. 由于不等式组恰有$$5$$个整数解，这$$5$$个整数解只能为$$15$$，$$16$$，$$17$$，$$18$$，$$19$$，因此$$14\\leqslant 3-2t ~\\textless{} ~15$$，解得$$-6\\textless t\\leqslant -\\frac{11}{2}$$． 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->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", "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" ]

[ "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． " ]

[ "1991年第2届希望杯初二竞赛第9题" ]

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

[ "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", "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" ]

[ "因为不论$$x$$为何实数，$${{x}^{2}}+\\textbar x\\textbar+1$$总是大于零的． 故选$$\\text{D}$$． " ]

[ "1994年第11届全国初中数学联赛竞赛第1题6分", "2016~2017学年广东广州天河区华南师范大学附属中学初二上学期期中第3题2分" ]

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

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

[ "∵$$0\\textless{}a\\textless{}1$$，∴$$a-\\frac{1}{a}\\textless{}0$$， ∴原式$$=\\sqrt{{{\\left( a-\\frac{1}{a} \\right)}^{2}}}\\div \\left( \\frac{a+1}{a} \\right)\\times \\frac{1}{1+a}$$ $$=\\left( \\frac{1}{a}-a \\right)\\times \\frac{a}{a+1}\\times \\frac{1}{1+a}$$ $$=\\frac{\\left( 1+a \\right)\\left( 1-a \\right)}{a}\\times \\frac{a}{a+1}\\times \\frac{1}{1+a}$$ $$=\\frac{1-a}{1+a}$$， 故答案为$$\\text{A}$$． " ]

[ "2013年第24届全国希望杯初二竞赛初赛第20题4分" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$7$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

[ "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->能力->运算能力", "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->算术基本定理" ]

[ "不大于$$20$$的正偶数的乘积是 $$2\\times 4\\times 6\\times 8\\times 10\\times 12\\times 14\\times 16\\times 18\\times 20$$ $$={{2}^{10}}\\times (1\\times 2\\times 3\\times 4\\times 5\\times 6\\times 7\\times 8\\times 9\\times 10)$$ $$={{2}^{10}}\\times (1\\times 2\\times 3\\times 4\\times 5\\times 6\\times 7\\times 8\\times 9\\times 10)$$ $$={{2}^{18}}\\times (3\\times 5\\times 3\\times 7\\times 9\\times 5)$$ $$={{2}^{18}}\\times {{3}^{4}}\\times {{5}^{2}}\\times 7$$． 设第一、二组中数的乘积分别是$$M$$，$$N$$， 则$$MN={{2}^{18}}\\times {{3}^{4}}\\times {{5}^{2}}\\times 7$$． 题意是要求最小的商，其实就是求最大的$$N$$， 易知$${{N}_{\\max }}={{2}^{9}}\\times {{3}^{2}}\\times 5$$， 故商为$$7$$． " ]

[ "初三上学期其它", "2003年第20届全国初中数学联赛竞赛第4题7分" ]

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

[ "课内体系->能力->运算能力", "课内体系->知识点->方程与不等式->其他方程->无理方程" ]

[ "已知等式可化为 $$\\sqrt{y}{{\\left( \\sqrt{x} \\right)}^{2}}+\\left[ {{\\left( \\sqrt{y} \\right)}^{2}}+\\sqrt{2003}\\sqrt{y}-\\sqrt{2003} \\right]\\sqrt{x}$$$$-\\sqrt{2003}\\left( \\sqrt{y}+\\sqrt{2003} \\right)=0$$， 故$$\\left( \\sqrt{y}\\sqrt{x}-\\sqrt{2003} \\right)\\left( \\sqrt{x}+\\sqrt{y}+\\sqrt{2003} \\right)=0$$． 又由题意， $$\\sqrt{x}+\\sqrt{y}+\\sqrt{2003}\\textgreater0$$， 则$$\\sqrt{y}\\sqrt{x}-\\sqrt{2003}=0\\Rightarrow xy=2003$$． 而$$2003$$为质数，且$$x$$、$$y$$为正整数，故$$\\left( x,y \\right)=\\left( 2003,1 \\right)$$，$$\\left( 1,2003 \\right)$$． 故选$$\\text{B}$$． " ]

[ "2010年第21届全国希望杯初一竞赛复赛第10题4分" ]

[ [ { "aoVal": "A", "content": "$$34$$ " } ], [ { "aoVal": "B", "content": "$$35$$ " } ], [ { "aoVal": "C", "content": "$$36$$ " } ], [ { "aoVal": "D", "content": "$$37$$ " } ] ]

[ "竞赛->知识点->数论->整除->算术基本定理", "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->点线面体" ]

[ "设正方体的每个面内的数分别为$${{x}_{1}}$$，$${{x}_{2}}$$，$${{x}_{3}}$$，$${{x}_{4}}$$，$${{x}_{5}}$$，$${{x}_{6}}$$， 则顶点上的数分别为$${{x}_{1}}{{x}_{2}}{{x}_{5}}$$，$${{x}_{2}}{{x}_{3}}{{x}_{5}}$$，$${{x}_{3}}{{x}_{4}}{{x}_{5}}$$，$${{x}_{4}}{{x}_{1}}{{x}_{5}}$$，$${{x}_{1}}{{x}_{2}}{{x}_{6}}$$，$${{x}_{2}}{{x}_{3}}{{x}_{6}}$$，$${{x}_{3}}{{x}_{4}}{{x}_{6}}$$，$${{x}_{4}}{{x}_{1}}{{x}_{6}}$$， 则有$$290={{x}_{1}}{{x}_{2}}{{x}_{5}}+{{x}_{2}}{{x}_{3}}{{x}_{5}}+{{x}_{3}}{{x}_{4}}{{x}_{5}}+{{x}_{4}}{{x}_{1}}{{x}_{5}}+{{x}_{1}}{{x}_{2}}{{x}_{6}}+{{x}_{2}}{{x}_{3}}{{x}_{6}}+{{x}_{3}}{{x}_{4}}{{x}_{6}}+{{x}_{4}}{{x}_{1}}{{x}_{6}}$$ $$=({{x}_{5}}+{{x}_{6}})({{x}_{1}}{{x}_{2}}+{{x}_{2}}{{x}_{3}}+{{x}_{3}}{{x}_{4}}+{{x}_{4}}{{x}_{1}})$$ $$=({{x}_{1}}+{{x}_{3}})({{x}_{2}}+{{x}_{4}})({{x}_{5}}+{{x}_{6}})$$． 由于$$290=2\\times 5\\times 29$$， 显然在乘积$$({{x}_{1}}+{{x}_{3}})({{x}_{2}}+{{x}_{4}})({{x}_{5}}+{{x}_{6}})$$中，一个因式等于$$2$$，一个因式等于$$5$$，一个因式等于$$29$$， 所以$${{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}+{{x}_{6}}=2+5+29=36$$． " ]

[ "1992年第3届全国希望杯初一竞赛复赛第2题" ]

[ [ { "aoVal": "A", "content": "正数 " } ], [ { "aoVal": "B", "content": "负数 " } ], [ { "aoVal": "C", "content": "奇数 " } ], [ { "aoVal": "D", "content": "偶数 " } ] ]

[ "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", "课内体系->知识点->数->有理数->相反数->相反数的性质" ]

[ "设该数为$$a$$，由题意$$-a$$为$$a$$的相反数，且有$${{a}^{3}}\\textless{}-a$$， ∴$${{a}^{3}}+a\\textless{}0$$，$$a({{a}^{2}}+1)\\textless{}0$$， 因为$${{a}^{2}}+1\\textgreater0$$，所以$$a\\textless{}0$$，即该数一定是负数，选$$\\text{B}$$． " ]

[ "1991年第2届希望杯初二竞赛第13题" ]

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

[ "课内体系->知识点->方程与不等式->其他方程->高次方程", "课内体系->方法->降次法" ]

[ "原方程可化为 $$\\left( 2{{x}^{5}}-20{{x}^{3}}+2x \\right)+\\left( {{x}^{4}}-10{{x}^{2}}+1 \\right)=0$$． 即$$\\left( 2x+1 \\right)\\left( {{x}^{4}}-10{{x}^{2}}+1 \\right)=0$$． ∴$$2x+1=0$$或$${{x}^{4}}-10{{x}^{2}}+1=0$$， 当$$2x+1=0$$时，解得$$x=-\\frac{1}{2}$$； 当$${{x}^{4}}-10{{x}^{2}}+1=0$$时，$${{x}^{2}}=\\frac{10+\\sqrt{{{10}^{2}}-4}}{2}=5+2\\sqrt{6}$$，或$${{x}^{2}}=\\frac{10-\\sqrt{{{10}^{2}}-4}}{2}=5-2\\sqrt{6}$$， ①当$${{x}^{2}}=5+2\\sqrt{6}$$，得$$x=\\sqrt{3}+\\sqrt{2}$$或$$x=-\\sqrt{3}-\\sqrt{2}$$， ②当$${{x}^{2}}=5-2\\sqrt{6}$$，得$$x=\\sqrt{3}-\\sqrt{2}$$或$$x=-\\sqrt{3}+\\sqrt{2}$$， 综上所述$$x$$可能为$$\\sqrt{3}+\\sqrt{2}$$， 故选$$\\text{C}$$． " ]

[ "1992年第3届希望杯初二竞赛第5题" ]

[ [ { "aoVal": "A", "content": "$$90{}^{}\\circ $$ " } ], [ { "aoVal": "B", "content": "$$105{}^{}\\circ $$ " } ], [ { "aoVal": "C", "content": "$$120{}^{}\\circ $$ " } ], [ { "aoVal": "D", "content": "$$150{}^{}\\circ $$ " } ] ]

[ "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", "课内体系->能力->运算能力" ]

[ "由$$\\angle A+\\angle B+\\angle C=180{}^{}\\circ $$，即$$\\left( \\theta -\\alpha \\right)+\\theta +\\left( \\theta +\\alpha \\right)=3\\theta =180{}^{}\\circ $$， 所以$$\\theta =\\angle ABC=60{}^{}\\circ $$， 因此$$\\frac{1}{2}\\left( \\angle BAC+\\angle BCA \\right)=\\frac{1}{2}\\left( 180{}^{}\\circ -60{}^{}\\circ \\right)=60{}^{}\\circ $$， 所以$$\\angle APC=180{}^{}\\circ -\\frac{1}{2}\\left( \\angle BAC+\\angle BCA \\right)=180{}^{}\\circ -60{}^{}\\circ =120{}^{}\\circ $$． 故应选$$\\text{C}$$． " ]

[ "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", "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" ]

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

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

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

[ "2015年第26届全国希望杯初一竞赛复赛第4题" ]

[ [ { "aoVal": "A", "content": "$$68$$ " } ], [ { "aoVal": "B", "content": "$$70$$ " } ], [ { "aoVal": "C", "content": "$$72$$ " } ], [ { "aoVal": "D", "content": "$$74$$ " } ] ]

[ "课内体系->知识点->方程与不等式->其他方程->三元一次方程组->解三元一次方程组", "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" ]

[ "设一班、二班、三班的人数分别为$$x$$、$$y$$、$$z$$． 由题意，得$$\\left { \\begin{array} ~\\&\\dfrac{x+y}{2}+z=45 \\dfrac{y+z}{2}+x=48 \\dfrac{x+z}{2}+y=47 \\end{array} \\right.$$， 三式相加，得$$2(x+y+z)=140$$， ∴$$x+y+z=70$$． 答：三个班的总人数为$$70$$人． " ]

[ "2015年第26届全国希望杯初一竞赛初赛第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->几何图形初步->命题与证明->猜想与证明->命题与证明的应用->区分真假命题" ]

[ "①正确； ②$$n$$为$$0$$时也满足要求，故错误； ③正确； ④如$$(-1)+(-2)=-3$$，$$-3\\textless{}-2$$且$$-3\\textless{}-1$$，故错误． 综上，正确的个数为两个． " ]

[ "1979年竞赛第13题" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$23$$ " } ] ]

[ "美国AMC8->Knowledge Point->Number and Operations->Integers->Mixed Basic Operations" ]

[ "$$10+20\\div 2+3=10+(20\\div 2)+3=10+10+3=23$$． 故选$$\\text{D}$$． " ]

[ "2012年第17届华杯赛初一竞赛初赛第4题" ]

[ [ { "aoVal": "A", "content": "$$-84$$ " } ], [ { "aoVal": "B", "content": "$$-89$$ " } ], [ { "aoVal": "C", "content": "$$-94$$ " } ], [ { "aoVal": "D", "content": "$$-99$$ " } ] ]

[ "竞赛->知识点->数与式->数的运算->有理数运算问题" ]

[ "要使算式的值尽可能小，应当减去一个较大的数，此时应使商尽可能小．如下式所示，所成的算式的值最小：$$10\\div 10\\times 10+10=-89$$． 故选$$\\text{B}$$． " ]

[ "2006年第17届希望杯初一竞赛初赛第8题4分" ]

[ [ { "aoVal": "A", "content": "$$10$$组 " } ], [ { "aoVal": "B", "content": "$$12$$组 " } ], [ { "aoVal": "C", "content": "$$15$$组 " } ], [ { "aoVal": "D", "content": "$$16$$组 " } ] ]

[ "竞赛->知识点->方程与不等式->一次方程->特殊方程" ]

[ "因为$$x$$，$$y$$ ，$$z$$均为正整数，且$$x+y+z=7$$， 所以$$1\\leqslant x\\leqslant 5$$． 下面分类讨论： 当$$x=1$$时，有$$5$$组解；当$$x=2$$时，有$$4$$组解； 当$$x=3$$时，有$$3$$组解；当$$x=4$$时，有$$2$$组解； 当$$x=5$$时，有$$1$$组解． 共计$$5+4+3+2+1=15$$（组）解． 故选$$\\text{C}$$． " ]

[ "2009年第20届希望杯初二竞赛第1试第4题" ]

[ [ { "aoVal": "A", "content": "$$1.414:1$$ " } ], [ { "aoVal": "B", "content": "$$2:1$$ " } ], [ { "aoVal": "C", "content": "$$1:0.618$$ " } ], [ { "aoVal": "D", "content": "$$1.732:1$$ " } ] ]

[ "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->矩形->矩形的性质" ]

[ "设$$A3$$型号的复印纸长为$$x$$，宽为$$y$$，对折后得到$$A4$$型号的复印纸长为$$y$$，宽为$$\\frac{x}{2}$$， 由题意，得$$\\frac{x}{y}=\\frac{y}{\\dfrac{x}{2}}$$， 即$${{x}^{2}}=2{{y}^{2}}$$， 所以$$x:y=\\sqrt{2}:1\\approx 1.414:1$$． 故选$$\\text{A}$$． " ]

[ "2010年第21届全国希望杯初一竞赛复赛第5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->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", "课内体系->能力->推理论证能力", "竞赛->知识点->数论->整除->整除的概念与基本性质" ]

[ "两位数的质数共有$$21$$个，它们的个位数字只有$$1$$，$$3$$，$$7$$，$$9$$； 差等于$$6$$的有：$$1$$和$$7$$；$$3$$和$$9$$；$$13$$和$$7$$，共三组； 平方数的个位数字相同的只有$$3$$和$$7$$，$$1$$和$$9$$这两组． 因此，符合题设条件的个位数字是$$3$$和$$7$$这一组． 故所求质数是$$23$$和$$17$$；$$43$$和$$37$$，$$53$$和$$47$$，$$73$$和$$67$$，共$$4$$组． ", "<p>平方之后个位数字相同的数有：$$1$$和$$9$$，$$2$$和$$8$$，$$3$$和$$7$$，$$4$$和$$6$$作为两位数质数的个位，不能是偶数，两个两位数相差$$6$$个位一定是$$7$$和$$3$$$$17$$和$$23$$，$$37$$和$$43$$，$$47$$和$$53$$，$$67$$和$$73$$，共$$4$$组．</p>" ]

[ "2000年第17届全国初中数学联赛竞赛第3题7分" ]

[ [ { "aoVal": "A", "content": "都不大于$$2$$ " } ], [ { "aoVal": "B", "content": "都不小于$$2$$ " } ], [ { "aoVal": "C", "content": "至少有一个大于$$2$$ " } ], [ { "aoVal": "D", "content": "至少有一个小于$$2$$ " } ] ]

[ "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->分式化简求值->分式化简求值-直接代入数值求值" ]

[ "令$$a=4$$，$$b=5$$，得到$$x=\\frac{13}{2}$$，$$y=\\frac{17}{5}$$，可排除$$\\text{A}$$，$$\\text{D}$$； 令$$a=1$$，$$b=2$$，得到$$x=5$$，$$y=1$$，可排除$$\\text{B}$$． 故只能选$$\\text{C}$$． " ]

[ "初一上学期单元测试《计算》第12题", "2006年第17届希望杯初二竞赛第1试第2题" ]

[ [ { "aoVal": "A", "content": "$$30$$ " } ], [ { "aoVal": "B", "content": "$$32$$ " } ], [ { "aoVal": "C", "content": "$$34$$ " } ], [ { "aoVal": "D", "content": "$$36$$ ~ " } ] ]

[ "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" ]

[ "∵$$441=7\\times 7\\times 3\\times 3$$， 又∵$$a$$，$$b$$，$$c$$，$$d$$互为不相等的正整数， ∴$$a$$，$$b$$，$$c$$，$$d$$为$$21$$，$$7$$，$$3$$，$$1$$， ∴$$a+b+c+d=32$$． " ]

[ "1997年第8届全国希望杯初一竞赛复赛第4题" ]

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

[ "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->分式方程的解" ]

[ "方程①的解$$x=1$$，将$$x=1$$代入方程②，方程②成立， ∴$$x=1$$也是方程②的解． 方程①和②是同解方程，而①与③显然不同解； ①的解代入④，④无意义． ∴$$\\text{B}$$、$$\\text{C}$$、$$\\text{D}$$都不正确，只有$$\\text{A}$$正确． 故选$$\\text{A}$$． " ]

[ "2010年第21届全国希望杯初一竞赛初赛第1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->正数和负数->正数、负数定义" ]

[ "因为$$a\\textless{}0$$， 所以$$\\left\\textbar{} a \\right\\textbar$$，$$-a$$，$${{a}^{2010}}$$，$$\\left\\textbar{} -a \\right\\textbar$$均为正数， 而$$\\left( \\frac{{{a}^{2}}}{a}-a \\right)=a-a=0$$，$$\\left( \\frac{{{a}^{2}}}{a}+a \\right)=a+a=2a$$， 所以只有$${{a}^{2009}}$$，$$\\left( \\frac{{{a}^{2}}}{a}+a \\right)$$为负数． 故负数的个数是$$2$$． " ]

[ "2019~2020学年湖北武汉期末", "1997年第8届全国希望杯初一竞赛复赛第4题" ]

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

[ "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->分式方程的解" ]

[ "方程①的解$$x=1$$，将$$x=1$$代入方程②，方程②成立， ∴$$x=1$$也是方程②的解． 方程①和②是同解方程，而①与③显然不同解； ①的解代入④，④无意义． ∴$$\\text{B}$$、$$\\text{C}$$、$$\\text{D}$$都不正确，只有$$\\text{A}$$正确． 故选$$\\text{A}$$． " ]

[ "2008年第19届希望杯初二竞赛第1试第5题" ]

[ [ { "aoVal": "A", "content": "$$x$$的最大值比$$y$$的最大值小 " } ], [ { "aoVal": "B", "content": "$$x$$的最小值比$$y$$的最小值小 " } ], [ { "aoVal": "C", "content": "$$x$$的最大值比$$y$$的最小值小 " } ], [ { "aoVal": "D", "content": "$$x$$的最小值比$$y$$的最大值大 " } ] ]

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

[ "$${{x}_{\\max }}=(-1)\\times 3+3\\times 2=3$$， $${{x}_{\\min }}=(-3)\\times 3+1\\times 2=-7$$， $${{y}_{\\max }}=9\\times 3-1\\times 2=25$$， $${{y}_{\\min }}=1\\times 3-9\\times 2=-15$$． ∴$$x$$的最大值比$$y$$的最大值小， 故选$$\\text{A}$$． " ]

[ "2013年第30届全国全国初中数学联赛竞赛第1题7分", "2013年全国全国初中数学联赛初一竞赛" ]

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

[ "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" ]

[ "$$4\\sqrt{3+2\\sqrt{2}}-\\sqrt{41+24\\sqrt{2}}$$ $$=4\\sqrt{{{(\\sqrt{2}+1)}^{2}}}-\\sqrt{{{(4\\sqrt{2}+3)}^{2}}}$$ $$=4(\\sqrt{2}+1)-(4\\sqrt{2}+3)$$ $$=1$$． " ]

[ "2014年第25届全国希望杯初一竞赛复赛第6题4分" ]

[ [ { "aoVal": "A", "content": "$$671$$ " } ], [ { "aoVal": "B", "content": "$$337$$ " } ], [ { "aoVal": "C", "content": "$$183$$ " } ], [ { "aoVal": "D", "content": "$$107$$ " } ] ]

[ "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->二元一次方程组的数字问题" ]

[ "设$$x$$为较大的数，$$y$$为较小的数， 则$$2013={{x}^{2}}-{{y}^{2}}=(x+y)(x-y)$$． ∵$$2013=1\\times 3\\times 11\\times 61$$， ∴满足题意的$$x+y$$是三位数或者四位数， 只能是$$\\left { \\begin{matrix}x+y=671 x-y=3 \\end{matrix} \\right.$$① 或$$\\left { \\begin{matrix}x+y=2013 x-y=1 \\end{matrix} \\right.$$② 或$$\\left { \\begin{matrix}x+y=183 x-y=11 \\end{matrix} \\right.$$③， 解①得$$\\left { \\begin{matrix}x=337 y=334 \\end{matrix} \\right.$$， 解②得$$\\left { \\begin{matrix}x=1007 y=1006 \\end{matrix} \\right.$$，（因为$$x$$，$$y$$都是四位数，故舍去） 解③得$$\\left { \\begin{matrix}x=97 y=86 \\end{matrix} \\right.$$，（因为$$x$$，$$y$$都是两位数，故舍去）． 综上，$$\\left { \\begin{matrix}x=337 y=334 \\end{matrix} \\right.$$， ∴这两个三位数中较大的一个是$$337$$． " ]

[ "2003年第14届希望杯初二竞赛第1试第2题" ]

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

[ "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" ]

[ "$${{x}^{4}}+5{{x}^{3}}y+{{x}^{2}}y+8{{x}^{2}}{{y}^{2}}+x{{y}^{2}}+5x{{y}^{3}}+{{y}^{4}}$$ $$=({{x}^{4}}+2{{x}^{2}}{{y}^{2}}+{{y}^{4}})+(5{{x}^{3}}y+5x{{y}^{3}})+6{{x}^{2}}{{y}^{2}}+{{x}^{2}}y+x{{y}^{2}}$$ $$={{({{x}^{2}}+{{y}^{2}})}^{2}}+5xy({{x}^{2}}+{{y}^{2}})+6{{x}^{2}}{{y}^{2}}+{{x}^{2}}y+x{{y}^{2}}$$ $$=({{x}^{2}}+{{y}^{2}}+2xy)({{x}^{2}}+{{y}^{2}}+3xy)+xy(x+y)$$ $$={{(x+y)}^{2}}({{x}^{2}}+{{y}^{2}}+3xy)+xy(x+y)$$ $$={{x}^{2}}+{{y}^{2}}+3xy-xy$$ $$={{x}^{2}}+{{y}^{2}}+2xy$$ $$={{(x+y)}^{2}}=1$$． " ]

[ "2015年第26届全国希望杯初一竞赛初赛第3题" ]

[ [ { "aoVal": "A", "content": "$$\\frac12a$$元 " } ], [ { "aoVal": "B", "content": "$$\\frac13a$$元 " } ], [ { "aoVal": "C", "content": "$$\\frac14a$$元 " } ], [ { "aoVal": "D", "content": "$$\\frac25a$$元 " } ] ]

[ "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" ]

[ "由题意，小明剩余的钱为$$(1-\\frac{1}{3})\\times (1-\\frac{1}{4})\\times (1-\\frac{1}{2})\\times a=\\frac{1}{4}a$$． " ]

[ "1996年第13届全国初中数学联赛竞赛第2题" ]

[ [ { "aoVal": "A", "content": "有$$1$$组 " } ], [ { "aoVal": "B", "content": "有$$2$$组 " } ], [ { "aoVal": "C", "content": "多于$$2$$组 " } ], [ { "aoVal": "D", "content": "不存在 " } ] ]

[ "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" ]

[ "将原式两边平方得到$${{a}^{2}}-4\\sqrt{2}=m+n-2\\sqrt{mn}$$，由于$$a$$，$$m$$，$$n$$都是正整数，所以$${{a}^{2}}=m+n$$，$$mn=8$$．又$$m\\textgreater n$$，仅当$$m=8$$，$$n=1$$时$$a$$为正整数．所以共有$$1$$组解．所以选$$\\text{A}$$． " ]

[ "2019~2020学年江苏苏州工业园区苏州星海实验中学初二下学期单元测试《分式》第40题2分", "2005年第16届希望杯初二竞赛初赛第8题4分" ]

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

[ "竞赛->知识点->方程与不等式->不等式->一次不等式组" ]

[ "因为$$a\\textless{}b\\textless{}c\\textless{}0$$， 所以$$0\\textless-c\\textless-b\\textless-a$$， 所以$$0\\textless-b-c\\textless-c-a\\textless-a-b$$， 于是$$\\frac{-c}{-a-b}\\textless\\frac{-b}{-c-a}\\textless\\frac{-a}{-b-c}$$， 所以$$\\frac{c}{a+b}\\textless{}\\frac{b}{c+a}\\textless{}\\frac{a}{b+c}$$． 故选$$\\text{D}$$． " ]

[ "2006年竞赛第4题6分" ]

[ [ { "aoVal": "A", "content": "$$2004$$ " } ], [ { "aoVal": "B", "content": "$$2005$$ " } ], [ { "aoVal": "C", "content": "$$2006$$ " } ], [ { "aoVal": "D", "content": "$$2007$$ " } ] ]

[ "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->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" ]

[ "根据题意，用剪刀沿不过顶点的直线剪成两部分时，每剪开一次，使得各部分的内角和增加$$360{}^{}\\circ $$．于是，剪过$$k$$次后，可得$$\\left( k+1 \\right)$$个多边形，这些多边形的内角和为$$\\left( k+1 \\right)\\times 360{}^{}\\circ $$． ∵这$$\\left( k+1 \\right)$$个多边形中有$$34$$个六十二边形，它们的内角和为$$34\\times \\left( 62-2 \\right)\\times 180{}^{}\\circ =34\\times 60\\times 180{}^{}\\circ $$，其余多边形有$$\\left( k+1 \\right)-34=k-33$$（个），而这些多边形的内角和不少于$$\\left( k-33 \\right)\\times 180{}^{}\\circ $$， ∴$$\\left( k+1 \\right)\\times 360{}^{}\\circ \\geqslant 34\\times 60\\times 180{}^{}\\circ +\\left( k-33 \\right)\\times 180{}^{}\\circ $$， 解得$$k\\geqslant 2005$$． 当我们按如下的方式剪$$2005$$刀时，可以得到符合条件的结论．先从正方形上剪下$$1$$个三角形，得到$$1$$个三角形和$$1$$个五边形；再在五边形上剪下$$1$$个三角形，得到$$2$$个三角形和$$1$$个六边形\\ldots\\ldots 如此下去，剪了$$58$$刀后，得到$$58$$个三角形和$$1$$个六十二边形．再取出$$33$$个三角形，在每个三角形上剪一刀，又可得到$$33$$个三角形和$$33$$个四边形，对这$$33$$个四边形，按上述正方形的剪法，再各剪$$58$$刀，便得到$$33$$个六十二边形和$$33\\times 58$$个三角形．于是共剪了$$55+33+33\\times 58=2005$$（刀）． " ]

[ "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", "2008年竞赛第2题6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", "初三上学期单元测试《概率初步》第23题", "2015~2016学年浙江温州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", "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" ]

[ [ { "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：$$17\\div 36=\\frac{17}{36}$$． " ]

[ "1998年第15届全国初中数学联赛竞赛第6题" ]

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

[ "竞赛->知识点->数论->同余->完全平方数", "课内体系->知识点->数->有理数->有理数运算巧解->完全平方非负性" ]

[ "整理得$$n^{2}-m^{2}=3995=5\\times17\\times47$$，$$(n-m)(n+m)=5\\times17\\times47$$， ∵对$$3995$$的任意整数分拆均可得到$$(m,n)$$，$$0\\textless m\\textless n\\textless1998$$， ∴$$\\begin{cases} n-m=5 n+m=17\\times47 \\end{cases}$$或$$\\begin{cases} n-m=17 n+m=5\\times47 \\end{cases}$$或$$ \\begin{cases} n-m=47 n+m=17\\times5 \\end{cases} $$或$$\\left { \\begin{array}{l} {m+n=5\\times17\\times47} {m-n=1} \\end{array}\\right. $$， ∴满足条件的整数对$$(m,n)$$共$$4$$个． 故答案为$$4$$． " ]

[ "2018~2019学年福建泉州南安市初二下学期期中第7题4分", "2018~2019学年山东潍坊高密市初二下学期期末第1题3分", "2019年广东惠州惠城区光正实验学校初二竞赛第7题3分" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$\\pm 3$$ " } ], [ { "aoVal": "D", "content": "$$3$$ " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->函数->一次函数->正比例函数->根据正比例函数定义求参数" ]

[ "∵$$y=(m+3){{x}^{{{m}^{2}}-8}}$$是正比例函数， ∴$${{m}^{2}}-8=1$$且$$m+3\\ne 0$$， 解得$$m=3$$． 故选$$\\text{D}$$． " ]

[ "初一下学期其它", "2006年第17届希望杯初二竞赛第2试第3题" ]

[ [ { "aoVal": "A", "content": "能作一个 " } ], [ { "aoVal": "B", "content": "能作两个 " } ], [ { "aoVal": "C", "content": "能作无数个 " } ], [ { "aoVal": "D", "content": "一个也不能作 " } ] ]

[ "课内体系->知识点->四边形->梯形->梯形的性质" ]

[ "由于梯形的两底之差$$a-c=3$$，以及梯形的两腰$$b=13$$，$$d=6$$不能构成三角形；故选$$\\text{D}$$． " ]

[ "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->知识点->式->整式的乘除->乘法公式->和与差的立方公式", "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->方程与不等式->不等式（组）->不等式->不等式的性质", "课内体系->能力->推理论证能力", "课内体系->能力->运算能力" ]

[ "∵$$a$$、$$b$$、$$c$$、$$d$$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$$\\sqrt[3]{7b+1}\\textgreater b+1\\sqrt[3]{7c+1}\\textgreater c+1\\sqrt[3]{7d+1}\\textgreater d+1$$， ∴$$p=\\sqrt[3]{7a+1}+\\sqrt[3]{7b+1}+\\sqrt[3]{7c+1}+\\sqrt[3]{7d+1}$$， $$\\textgreater a+1+b+1+c+1+d+1$$， $$=a+b+c+d+4=5$$， ∴$$p\\textgreater5$$． " ]

[ "2007年第18届希望杯初二竞赛第1试第9题" ]

[ [ { "aoVal": "A", "content": "$$9$$或$$18$$ " } ], [ { "aoVal": "B", "content": "$$12$$或$$15$$ " } ], [ { "aoVal": "C", "content": "$$9$$或$$15$$或$$18$$ " } ], [ { "aoVal": "D", "content": "$$9$$或$$12$$或$$15$$或$$18$$ " } ] ]

[ "竞赛->知识点->三角形->三角形基础" ]

[ "由$${{x}^{2}}-9x+18=\\left( x-3 \\right)\\left( x-6 \\right)$$知，使代数式$${{x}^{2}}-9x+18$$的值为$$0$$的$$x$$的值是$$3$$或$$6$$，于是三角形三边的长有四种可能情况： ①三边长分别是$$6$$，$$6$$，$$6$$，则三角形的周长为$$18$$； ②三边长分别是$$6$$，$$6$$，$$3$$，则三角形的周长为$$15$$； ③三边长分别是$$3$$，$$3$$，$$3$$，则三角形的周长为$$9$$； ④三边长分别是$$6$$，$$3$$，$$3$$，则不能构成三角形． 综上知，三角形的周长是$$9$$，$$15$$或$$18$$． 故选$$\\text{C}$$． " ]

[ "1996年第7届全国希望杯初一竞赛初赛第8题" ]

[ [ { "aoVal": "A", "content": "甲 " } ], [ { "aoVal": "B", "content": "乙 " } ], [ { "aoVal": "C", "content": "丙 " } ], [ { "aoVal": "D", "content": "丁 " } ] ]

[ "课内体系->知识点->几何图形初步->角->角的定义和分类->角的分类" ]

[ "∵$$90{}^{}\\circ \\textless{}\\alpha \\textless{}180{}^{}\\circ $$，$$90{}^{}\\circ \\textless{}\\beta \\textless{}180{}^{}\\circ $$，∴$$180{}^{}\\circ \\textless{}\\alpha +\\beta \\textless{}360{}^{}\\circ $$， ∴$$30{}^{}\\circ \\textless{}\\frac{1}{6}(\\alpha +\\beta )\\textless{}60{}^{}\\circ $$，可见$$\\frac{1}{6}(\\alpha +\\beta )$$应是$$50{}^{}\\circ $$． 故甲计算正确． " ]

[ "1978年Math League竞赛(8年级[.org])第15题", "1978年竞赛第15题" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$2$$，$$3$$和$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "没有选项 " } ] ]

[ "美国AMC8->Knowledge Point->Algebra and Sequences->Equations->Inequality", "课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式->一元一次不等式的整数解", "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" ]

[ "The inequality $$3x -1\\textless{} 11$$ is satisfied by only $$2$$ \\& $$3$$， Choice $$\\text{A}$$ is incomplete， choice $$\\text{B}$$ includes $$4$$， \\& choice $$\\text{C}$$ is incorrect． 在数$$2$$，$$3$$，$$4$$，$$5$$中，哪一个是（唯一）满足不等式$$3x-1\\textless11$$的． $$\\text{A}$$．$$2$$ ~ ~ ~ $$\\text{B}$$．$$2$$，$$3$$和$$4$$ ~ ~ ~ $$\\text{C}$$．$$5$$ ~ ~ ~ $$\\text{D}$$．没有选项 满足不等式$$3x-1\\textless{}11$$只有$$2$$和$$3$$．选项$$\\text{A}$$不完整，选项$$\\text{B}$$包含$$4$$，选项$$\\text{C}$$不正确． 故选$$\\text{D}$$． " ]

[ "1992年第3届希望杯初二竞赛第3题" ]

[ [ { "aoVal": "A", "content": "$$1200000-254000$$ " } ], [ { "aoVal": "B", "content": "$$1020000-250400$$ " } ], [ { "aoVal": "C", "content": "$$1200000-250400$$ " } ], [ { "aoVal": "D", "content": "$$1020000-254000$$ " } ], [ { "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 " ]

[ "2002年第13届希望杯初一竞赛第1试第6题" ]

[ [ { "aoVal": "A", "content": "$$25 \\%$$ " } ], [ { "aoVal": "B", "content": "$$20 \\%$$ " } ], [ { "aoVal": "C", "content": "$$8 \\%$$ " } ], [ { "aoVal": "D", "content": "$$12 \\%$$ " } ] ]

[ "课内体系->能力->运算能力", "课内体系->知识点->数->有理数->有理数与实际问题" ]

[ "由题意可得，$$5$$月份销售该品牌的$$\\text{VCD}$$机可获利 $$=\\frac{625 \\times(1-20 \\%) \\times(1+8 \\%)-500}{500}$$ $$=\\frac{625 \\times 0.8 \\times 1.08}{500}-1$$ $$=8 \\%$$． 故选$$\\text{C}$$． " ]

[ "初一下学期单元测试《质数与合数》第2题", "2005年第16届希望杯初二竞赛初赛第3题4分" ]

[ [ { "aoVal": "A", "content": "$$1$$个 " } ], [ { "aoVal": "B", "content": "$$3$$个 " } ], [ { "aoVal": "C", "content": "$$5$$个 " } ], [ { "aoVal": "D", "content": "$$6$$个~ " } ] ]

[ "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" ]

[ "满足条件的两位数两个数字之差为$$1$$， 这样的两位数有$$12$$、$$23$$、$$34$$、$$45$$、$$56$$、$$67$$、$$78$$、$$89$$， 其中质数有$$23$$，$$67$$、$$89$$． " ]

[ "1996年第7届希望杯初二竞赛第7题" ]

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

[ "竞赛->知识点->数与式->分式->倒数通分" ]

[ "∵$$\\frac{1}{a}:\\frac{1}{b}:\\frac{1}{c}=2:3:4$$， ∴$$a:b:c=\\frac{1}{2}:\\frac{1}{3}:\\frac{1}{4}=6:4:3$$． 故选$$\\text{B}$$． " ]

[ "2009年第20届希望杯初二竞赛第1试第6题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{2}{3}\\textless{}x\\textless{}2$$ " } ], [ { "aoVal": "B", "content": "$$x\\textgreater\\frac{2}{3}$$或$$x\\textless{}-2$$ " } ], [ { "aoVal": "C", "content": "$$-2\\textless{}x\\textless{}2$$且$$x\\ne \\frac{2}{3}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{2}{3}\\textless{}x\\textless{}2$$或$$x\\textless{}-2$$ " } ] ]

[ "竞赛->知识点->数与式->分式->分式的基本运算" ]

[ "因为分式$$\\frac{\\textbar x\\textbar-2}{3x-2}$$的值为负，所以$$\\textbar x\\textbar-2$$和$$3x-2$$异号． 当$$3x-2\\textgreater0$$，即$$x\\textgreater\\frac{2}{3}$$时，应当有 $$\\left\\textbar{} x \\right\\textbar-2\\textless{}0$$，$$\\left\\textbar{} x \\right\\textbar\\textless{}2$$，解得$$-2\\textless{}x\\textless{} 2$$， 所以$$\\frac{2}{3}\\textless{}x\\textless{}2$$ ． 当$$3x-2\\textless{}0$$，即$$x\\textless{}\\frac{2}{3}$$时，应当有 $$\\left\\textbar{} x \\right\\textbar-2\\textgreater0$$，$$\\left\\textbar{} x \\right\\textbar\\textgreater2$$，解得$$x\\textgreater2$$或$$x\\textless{}-2$$， 所以$$x\\textless{}-2$$． 综上可知，$$x$$的取值范围是$$\\frac{2}{3}\\textless{}x\\textless{}2$$或$$x\\textless{}-2$$． 故选$$\\text{D}$$． " ]

[ "1991年第2届全国希望杯初一竞赛初赛第5题" ]

[ [ { "aoVal": "A", "content": "$$(-13579)+0.2468$$ " } ], [ { "aoVal": "B", "content": "$$(-13579)+\\frac{1}{2468}$$ " } ], [ { "aoVal": "C", "content": "$$(-13579)\\times \\frac{1}{2468}$$ " } ], [ { "aoVal": "D", "content": "$$(-13579)\\div \\frac{1}{2468}$$ " } ] ]

[ "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->有理数乘方->含乘方的有理数混合运算" ]

[ "运算结果对负数来说绝对值越小其值越大． 易见$$(-13579)\\times \\frac{1}{2468}$$的绝对值最小，所以其值最大． 选$$\\text{C}$$． " ]

[ "1997年第8届希望杯初二竞赛第2试第6题", "初一下学期其它" ]

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

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

[ "因为$${{m}^{2}}+m-1=0$$， 所以$${{m}^{2}}+m=1$$， 所以$${{m}^{3}}+2{{m}^{2}}-1997$$ $$=m({{m}^{2}}+m)+{{m}^{2}}-1997$$ $$=m+{{m}^{2}}-1997$$ $$=1-1997$$ $$=-1996$$． " ]

[ "1990年第1届全国希望杯初一竞赛初赛第9题" ]

[ [ { "aoVal": "A", "content": "一样多 " } ], [ { "aoVal": "B", "content": "多了 " } ], [ { "aoVal": "C", "content": "少了 " } ], [ { "aoVal": "D", "content": "多少都可能 " } ] ]

[ "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" ]

[ "设杯中原有水量为$$a$$，依题意可得， 第二天杯中水量为$$a\\times (1-10 \\%)=0.9a$$； 第三天杯中水量为$$(0.9a)\\times (1+10 \\%)=0.9\\times 1.1\\times a$$； 第三天杯中水量与第一天杯中水量之比为$$\\frac{0.9\\times 1.1\\times a}{a}=0.99\\textless1$$． 所以第三天杯中水量比第一天杯中水量少了，选$$\\text{C}$$． " ]

[ "2015年第32届全国全国初中数学联赛竞赛B卷第4题7分" ]

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

[ "课内体系->方法->换元法", "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->由一元二次方程根的情况确定参数", "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" ]

[ "设$$x-y=t$$，则$$x=y+t$$， 代入$${{x}^{2}}+xy+{{y}^{2}}=3$$， 得$${{(y+t)}^{2}}+(y+t)y+{{y}^{2}}=3$$， 整理，得$$3{{y}^{2}}+3ty+{{t}^{2}}-3=0$$． 由判别式$$\\Delta ={{(3t)}^{2}}-12({{t}^{2}}-3)\\geqslant 0$$， 得$$-2\\sqrt{3}\\leqslant t\\leqslant 2\\sqrt{3}$$， ∴$${{(x-y)}^{2}}={{t}^{2}}\\leqslant 12$$， 即$${{(x-y)}^{2}}$$的最大值为$$12$$． " ]

[ "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", "2001年第18届全国初中数学联赛竞赛第1题", "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", "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" ]

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

[ "课内体系->能力->运算能力", "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->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" ]

[ "\\textbf{（知识点：根式恒等变形）} 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$$\\text{B}$$． " ]

[ "2014年第31届全国全国初中数学联赛竞赛第4题7分" ]

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

[ "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" ]

[ "若取出的$$3$$张卡片上的数字互不相同，有$$2\\times 2\\times 2=8$$种取法； 若取出的$$3$$张卡片上的数字有相同的，有$$3\\times 4=12$$种取法． 所以，从$$6$$张不同的卡片中取出$$3$$张，共有$$812=20$$种取法． 要使得三个数字可以构成三角形的三边长， 只可能是：$$(2,4,4)$$，$$(4,4,6)$$，$$(2,6,6)$$，$$(4,6,6)$$， 由于不同的卡片上所写数字有重复， 所以，取出的$$3$$张卡片上所写的数字可以作为三角形的三边长的情况共有$$4\\times 2=8$$种． 因此，所求概率为$$\\frac{8}{20}=\\frac{2}{5}$$． " ]

[ "1990年第1届全国希望杯初一竞赛初赛第3题" ]

[ [ { "aoVal": "A", "content": "有最小的自然数 " } ], [ { "aoVal": "B", "content": "没有最小的正有理数 " } ], [ { "aoVal": "C", "content": "没有最大的负整数 " } ], [ { "aoVal": "D", "content": "没有最大的非负数 " } ] ]

[ "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" ]

[ "$$0$$是最小的自然数，$$\\text{A}$$正确． 可以找到正有理数的无限序列$$1$$，$$\\frac{1}{2}$$，$$\\frac{1}{3}$$，$$\\cdots $$，$$\\frac{1}{n}$$，$$\\cdots $$，没有最小的正有理数，$$\\text{B}$$正确． ``没有最大的负整数''的说法不正确，最大的负整数为$$-1$$． 写出自然数列，$$0$$，$$1$$，$$2$$，$$3$$，$$\\ldots $$，$$n$$，$$\\ldots $$，易知无最大非负数，$$\\text{D}$$正确． 所以不正确的说法应选$$\\text{C}$$． " ]

[ "2013年第24届全国希望杯初二竞赛复赛第3题4分" ]

[ [ { "aoVal": "A", "content": "$$12532$$ " } ], [ { "aoVal": "B", "content": "$$12533$$ " } ], [ { "aoVal": "C", "content": "$$12534$$ " } ], [ { "aoVal": "D", "content": "$$12535$$ " } ] ]

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

[ "由题意，可得 $${{(a-b)}^{2}}+{{(a+b)}^{2}}+{{(b-c)}^{2}}+{{(b+c)}^{2}}+{{(a-c)}^{2}}+{{(a+c)}^{2}}$$ $$={{a}^{2}}-2ab+{{b}^{2}}+{{a}^{2}}+2ab+{{b}^{2}}+{{b}^{2}}-2bc+{{c}^{2}}+{{b}^{2}}+2bc+{{c}^{2}}+{{a}^{2}}-2ac+{{c}^{2}}+{{a}^{2}}+2ac+{{c}^{2}}$$ $$=4({{a}^{2}}+{{b}^{2}}+{{c}^{2}})$$， 所以$${{a}^{2}}+{{b}^{2}}+{{c}^{2}}$$ $$=\\frac{{{42}^{2}}+{{45}^{2}}+{{64}^{2}}+{{87}^{2}}+{{109}^{2}}+{{151}^{2}}}{4}$$ $$=12534$$． " ]

[ "1994年第5届希望杯初二竞赛第5题" ]

[ [ { "aoVal": "A", "content": "①②③ " } ], [ { "aoVal": "B", "content": "①②④ " } ], [ { "aoVal": "C", "content": "②③④ " } ], [ { "aoVal": "D", "content": "①③④ " } ] ]

[ "课内体系->能力->抽象概括能力", "课内体系->知识点->数->实数->立方根->开立方" ]

[ "负数有立方根，$$0$$的立方根是$$0$$，又$$-1$$的立方根也是$$-1$$，所以错误命题是①②④． 应选$$\\text{B}$$． " ]

[ "2006年第17届希望杯初二竞赛第1试第5题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{9}{19}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{10}{19}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{11}{21}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{10}{21}$$ " } ] ]

[ "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", "竞赛->知识点->方程与不等式->方程应用" ]

[ "设初一年级男生有$$m$$人，女生有$$n$$人，则初二年级男生有$$n$$人，女生有$$m$$人；初三年级的学生共有$$\\frac{4}{5}(m+n)$$人；三个年级共有学生$$\\frac{14}{5}(m+n)$$人． 设初三年级男生有$$x$$人，则$$x=\\frac{1}{4}(m+n+x)$$， 解得$$x=\\frac{1}{3}(m+n)$$， 所以，初三年级的女生人数为$$\\frac{4}{5}(m+n)-\\frac{1}{3}(m+n)=\\frac{7}{15}(m+n)$$， 故全校女生人数为$$\\frac{22}{15}(m+n)$$人，占三个年级学生人数的$$\\frac{11}{21}$$． 故选$$\\text{C}$$． " ]

[ "2013年山东青岛黄岛区山东省青岛第九中学自主招生", "2012年竞赛第2题4分" ]

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

[ "竞赛->知识点->函数->反比例函数->反比例函数与一次函数综合" ]

[ "由题设知，$$-2=a\\cdot (-3)$$，$$(-3)\\cdot (-2)=b$$， 所以$$a=\\frac{2}{3}$$，$$b=6$$， 解方程组$$\\begin{cases}y=\\dfrac{2}{3}x y=\\dfrac{6}{x} \\end{cases}$$， 得$$\\begin{cases}x=-3 y=-2 \\end{cases}$$， 或$$\\begin{cases}x=3 y=2 \\end{cases}$$， 所以另一个交点的坐标为$$(3,2)$$． " ]

[ "2020年第22届浙江宁波余姚市余姚市实验学校初三竞赛（实验杯）第11题4分" ]

[ [ { "aoVal": "A", "content": "有最大值，又有最小值 " } ], [ { "aoVal": "B", "content": "有最大值，没有最小值 " } ], [ { "aoVal": "C", "content": "有最小值，没有最大值 " } ], [ { "aoVal": "D", "content": "没有最大值，也没有最小值 " } ] ]

[ "课内体系->能力->分析和解决问题能力", "课内体系->知识点->式->整式的加减->整式加减化简求值->整式加减化简求值-条件化简求值", "课内体系->知识点->式->整式的乘除->乘法公式->配方求最值" ]

[ "∵$$a-b^{2}=b-a^{2}$$， ∴$$a^{2}-b^{2}=b-a$$， $$(a+b)(a-b)=b-a$$ $$(a+b+1)(a-b)=0$$， 又∵$$a\\ne b$$， ∴$$a+b+1=0$$， ∴$$a=-b-1$$， ∴$$x=a-b^{2}$$ $$=-b-1-b^{2}$$ $$=-\\left(b+\\frac{1}{2} \\right)^{2}-\\frac{3}{4}$$， 当$$b=-\\frac{1}{2}$$时，$$x$$取最大值$$-\\frac{3}{4}$$， 但此时$$a=-\\left(-\\frac{1}{2}\\right)-1=-\\frac{1}{2}=b$$， 故最大值取不到，也没有最小值， 故选$$\\text{D}$$． " ]

[ "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": "$$\\frac{8}{7}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{16}{7}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{32}{7}$$ " } ] ]

[ "竞赛->知识点->数与式->因式分解->因式分解：添项、拆项", "课内体系->知识点->方程与不等式->不等式（组）->其它不等式->均值不等式" ]

[ "不要搜题 " ]

[ "2010年全美数学竞赛（AMC）竞赛" ]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$8$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ], [ { "aoVal": "E", "content": "$$11$$ " } ] ]

[ "美国AMC8->Knowledge Point->Geometry->Shapes with Straight Sides->Geometric Theorem", "课内体系->知识点->三角形" ]

[ "某三角形三边长为连续正整数．最短边占周长$$30 \\%$$．则最长边长度是？ 设$$n$$，$$n+1$$，和$$n+2$$是三角形三边的长度．那么三角形的周长是$$n+(n+1)+(n+2)=3n+3$$利用最小边的长度是周长的$$30 \\%$$这个事实，可以得出这样的结论： $$n=0.3(3n+3)\\Rightarrow 0.9n+0.9\\Rightarrow 0.1n=0.9\\Rightarrow n=9$$．最长边是$$n+2=11$$． 故选$$\\text{E}$$． 由于最短边的长度是整数，等于周界的$$\\frac{3}{10}$$，因此周界必须是$$10$$的倍数．将前面的两个整数添加到每个答案选择中，我们可以看到$$11+10+9=30$$． 故选$$\\text{E}$$． Let $$n$$, $$n+1$$, and $$n+2$$ be the lengths of the sides of the triangle. Then the perimeter of the triangle is $$n+\\left( n+1 \\right)+\\left( n+2 \\right)=3n+3$$. Usin g the fact that the length of the smallest side is $$30 \\%$$ of the perimeter, it follows that: $$n=0.3\\left( 3n +3 \\right)\\Rightarrow n=0.9n+0.9\\Rightarrow 0.1n=0.9\\Rightarrow n=9$$. The longest side is then $$n+2-11$$. Thus, answer choice $$\\left( \\text{E} \\right)11$$ is correct. Since the length of the shortest side is a whole number and is equal to $$\\frac{3}{10}$$ of the perimeter, it follows that the perimeter must be a multiple of $$10$$. Adding the two previous integers to each answer choice, we see that $$11+10+9=30$$. Thus, answer choice $$\\left( \\text{E} \\right)11$$ is correct. " ]

[ "2019年第2届浙江宁波余姚市余姚市实验学校初二竞赛第8题5分" ]

[ [ { "aoVal": "A", "content": "$$15$$ " } ], [ { "aoVal": "B", "content": "$$17$$ " } ], [ { "aoVal": "C", "content": "$$19$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式组->一元一次不等式组的整数解", "课内体系->知识点->方程与不等式->不等式（组）->一元一次不等式组->解一元一次不等式组" ]

[ "解不等式组$$\\begin{cases}\\dfrac{5x+5}{6}-x\\geqslant -\\dfrac{5}{2}① \\dfrac{2x+5}{3}\\textless{}nx② \\end{cases}$$， 由①得，$$5x+5-6x\\geqslant -15$$， $$5x-6x\\geqslant -15-5$$， $$-x\\geqslant -20$$， $$x\\leqslant 20$$， 由②得$$2x+5\\textless{}3nx$$， $$2x-3nx\\textless{}-5$$， $$\\left( 2-3n \\right)x\\textless{}-5$$， ∵$$n$$为正整数， ∴$$n\\geqslant 1$$， ∴$$2-3n\\leqslant -1$$， ∴解②，得$$x\\textgreater\\frac{5}{3n-2}$$， ∴原不等式组的解集为$$\\frac{5}{3n-2}\\textless{}x\\leqslant 20$$， ∴当$$n=1$$时，$$5\\textless{}x\\leqslant 20$$，有$$15$$个整数解， 当$$n=2$$时，$$\\frac{5}{4}\\textless{}x\\leqslant 20$$，有$$19$$个整数解， 当$$n=3$$时，$$\\frac{5}{7}\\textless{}x\\leqslant 20$$，有$$20$$个整数解， ∴原不等式组不可能有$$17$$个整数解． 故选$$\\text{B}$$． " ]

[ "2011年第28届全国全国初中数学联赛竞赛第6题7分" ]

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

[ "课内体系->知识点->式->分式->分式化简求值->分式化简求值-倒数法化简求值" ]

[ "法1：由通分然后取倒数得， $$\\frac{xy+xz}{x+y+z}=2$$，$$\\frac{xy+yz}{x+y+z}=3$$，$$\\frac{zx+yz}{x+y+z}=4$$， $$\\therefore \\frac{2}{x}+\\frac{3}{y}+\\frac{4}{z}=\\frac{xy+zx}{x+y+z}\\cdot \\frac{1}{x}+\\frac{xy+yz}{x+y+z}\\cdot \\frac{1}{y}+\\frac{zx+zy}{x+y+z}\\cdot \\frac{1}{z}$$ $$=\\frac{x+y+x+z+y+z}{x+y+z}=2$$． 法2：由已知等式得$$\\frac{xy+zx}{x+y+z}=2$$，$$\\frac{yz+xy}{x+y+z}=3$$，$$\\frac{zx+yz}{x+y+z}=4$$， 所以$$\\frac{xy+yz+zx}{x+y+z}=\\frac{9}{2}$$． 于是，$$\\frac{yz}{x+y+z}=\\frac{5}{2}$$，$$\\frac{zx}{x+y+z}=\\frac{3}{2}$$，$$\\frac{xy}{x+y+z}=\\frac{1}{2}$$． 所以$$\\frac{y}{x}=\\frac{5}{3}$$，$$\\frac{z}{y}=3$$，$$\\frac{y}{x}=\\frac{5}{3}$$， 即$$z=3y=5x$$． 代入$$\\frac{1}{x}+\\frac{1}{y+z}=\\frac{1}{2}$$，得$$\\frac{1}{x}+\\frac{1}{\\dfrac{5}{3}x+5x}=\\frac{1}{2}$$， 解得$$x=\\frac{23}{10}$$． 所以$$\\frac{2}{x}+\\frac{3}{y}+\\frac{4}{z}=\\frac{2}{x}+\\frac{3}{\\dfrac{5}{3}x}+\\dfrac{4}{5x}=\\frac{23}{5x}=2$$． " ]

[ "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", "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" ]

[ [ { "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：$$\\text{C}$$． " ]

[ "1991年第2届希望杯初二竞赛第7题" ]

[ [ { "aoVal": "A", "content": "必有一个钝角 " } ], [ { "aoVal": "B", "content": "必有一个锐角 " } ], [ { "aoVal": "C", "content": "必有一个不是钝角 " } ], [ { "aoVal": "D", "content": "必有两个锐角 " } ] ]

[ "课内体系->知识点->几何图形初步->角->角的定义和分类->角的分类" ]

[ "相交时两条直线可能互相垂直（垂直是一种特殊的相交）， ∴$$\\text{A}$$、$$\\text{B}$$、$$\\text{D}$$均在垂直的情况下不成立． ", "<p>$$\\text{A}$$．两个$$90{}^\\circ $$，故$$\\text{A}$$错误；</p>\n<p>$$\\text{B}$$．两个$$90{}^\\circ $$，故$$\\text{B}$$错误；</p>\n<p>$$\\text{C}$$．一钝角、一锐角或两个直角，故$$\\text{C}$$正确；</p>\n<p>$$\\text{D}$$．和小于$$180{}^\\circ $$，故$$\\text{D}$$错误．</p>\n<p>故选$$\\text{C}$$．</p>" ]

[ "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->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$$a\\textless{}b\\textless{}0$$ $$\\therefore a+b=-\\sqrt{6ab}$$，$$a-b=-\\sqrt{2ab}\\therefore \\frac{a+b}{a-b}=\\sqrt{3}$$ " ]

[ "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->知识点->式->整式的乘除->乘法公式->完全平方公式的计算-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$$7$$． " ]