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八年级数学填空题一般
题目
如果两个分式MMNN的和为常数kk,且kk为正整数,则称MMNN互为"和整分式",常数kk称为"和整值".如分式M=xx+1M=\frac{x}{x+1},N=1x+1N=\frac{1}{x+1},M+N=x+1x+1=1M+N=\frac{x+1}{x+1}=1,则MMNN互为"和整分式","和整值"k=1k=1.
(1)(1)已知分式A=x7x2A=\frac{x-7}{x-2},B=2x+1x2B=\frac{2x+1}{x-2},判断AABB是否互为"和整分式",若不是,请说明理由;若是,请求出"和整值"kk
(2)(2)已知分式C=3x4x2C=\frac{3x-4}{x-2},D=Gx24D=\frac{G}{x^2-4},CCDD互为"和整分式",且"和整值"k=3k=3.
①求GG
②若xx为正整数,分式DD的值也为正整数.则xx的值为______.
知识点:解二元一次方程组——代入消元法、解二元一次方程组——加减消元法、解二元一次方程组章节:未标注

答案与解析

答案

(1)A+B=x7x2+2x+1x2=x7+2x+1x2=3(x2)x2=3\left(1\right)\because A+B=\frac{x-7}{x-2}+\frac{2x+1}{x-2}=\frac{x-7+2x+1}{x-2}=\frac{3(x-2)}{x-2}=3
\therefore分式AA与分式BB是互为“和整分式”,“和整值”k=3k=3
(2)(2)\because分式C=3x4x2C=\frac{3x-4}{x-2}D=Gx24D=\frac{G}{x^2-4}CCDD互为“和整分式”,且“和整值”k=3k=3
C+D=3x4x2+Gx24=3\therefore C+D=\frac{3x-4}{x-2}+\frac{G}{x^{2}-4}=3
两边都乘以(x+2)(x2)\left(x+2\right)\left(x-2\right)得,
(3x4)(x+2)+G=3(x24)(3x-4)\left(x+2\right)+G=3(x^{2}-4)
G=3(x24)(3x4)(x+2)\therefore G=3(x^{2}-4)-\left(3x-4\right)\left(x+2\right)
=3x2123x26x+4x+8=3x^{2}-12-3x^{2}-6x+4x+8
=2x4=-2x-4
D=2x4x24=22x\because D=\frac{-2x-4}{x^{2}-4}=\frac{2}{2-x},而xx为正整数,分式DD的值也为正整数,
x=1\therefore x=1
故答案为:11.

解析

(1)A+B=x7x2+2x+1x2=x7+2x+1x2=3(x2)x2=3\left(1\right)\because A+B=\frac{x-7}{x-2}+\frac{2x+1}{x-2}=\frac{x-7+2x+1}{x-2}=\frac{3(x-2)}{x-2}=3
\therefore分式AA与分式BB是互为“和整分式”,“和整值”k=3k=3
(2)(2)\because分式C=3x4x2C=\frac{3x-4}{x-2}D=Gx24D=\frac{G}{x^2-4}CCDD互为“和整分式”,且“和整值”k=3k=3
C+D=3x4x2+Gx24=3\therefore C+D=\frac{3x-4}{x-2}+\frac{G}{x^{2}-4}=3
两边都乘以(x+2)(x2)\left(x+2\right)\left(x-2\right)得,
(3x4)(x+2)+G=3(x24)(3x-4)\left(x+2\right)+G=3(x^{2}-4)
G=3(x24)(3x4)(x+2)\therefore G=3(x^{2}-4)-\left(3x-4\right)\left(x+2\right)
=3x2123x26x+4x+8=3x^{2}-12-3x^{2}-6x+4x+8
=2x4=-2x-4
D=2x4x24=22x\because D=\frac{-2x-4}{x^{2}-4}=\frac{2}{2-x},而xx为正整数,分式DD的值也为正整数,
x=1\therefore x=1
故答案为:11.

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