《2025年通城学典暑期升级训练延边大学出版社八年级数学华师大版》

5. (东营中考)先化简,再求值:$(\frac {1}{x-y}-\frac {1}{x+y})÷$$\frac {2y}{x^{2}+2xy+y^{2}}$,其中$x= 3,y= 2.$

答案： 原式$=\left[\frac{x+y}{(x+y)(x-y)}-\frac{x-y}{(x+y)(x-y)}\right]\cdot\frac{(x+y)^2}{2y}=\frac{2y}{(x+y)(x-y)}\cdot\frac{(x+y)^2}{2y}=\frac{x+y}{x-y}$. 当$x=3,y=2$时,原式$=\frac{3+2}{3-2}=5$.

6. (营口中考)先化简,再求值:$(a+1-\frac {5+2a}{a+1})÷\frac {a^{2}+4a+4}{a+1}$,其中$a= \sqrt {9}+|-2|-(\frac {1}{2})^{-1}.$

答案： 原式$=\frac{(a+1)^2-(5+2a)}{a+1}\cdot\frac{a+1}{(a+2)^2}=\frac{a^2+2a+1-5-2a}{a+1}\cdot\frac{a+1}{(a+2)^2}=\frac{a^2-4}{a+1}\cdot\frac{a+1}{(a+2)^2}=\frac{(a+2)(a-2)}{a+1}\cdot\frac{a+1}{(a+2)^2}=\frac{a-2}{a+2}$.$\because a=\sqrt{9}+|-2|-\left(\frac{1}{2}\right)^{-1}=3+2-2=3$,$\therefore$原式$=\frac{3-2}{3+2}=\frac{1}{5}$.

7. (黄石中考)先化简,再求值:$(1+\frac {2}{a+1})÷$$\frac {a^{2}+6a+9}{a+1}$,从-3、-1、2中选择合适的a
的值代入求值.

答案： 原式$=\frac{a+3}{a+1}÷\frac{(a+3)^2}{a+1}=\frac{a+3}{a+1}\cdot\frac{a+1}{(a+3)^2}=\frac{1}{a+3}$.要使分式有意义,则$a$不能为$-1$、$-3$.$\therefore a=2$.原式$=\frac{1}{2+3}=\frac{1}{5}$.

8. 先化简$(\frac {x^{2}-1}{x^{2}-2x+1}+\frac {1}{1-x})÷\frac {x^{2}}{x-1}$,若x的
取值范围是$-1≤x≤1$,且为整数,求该式
的值.

答案： 原式$=\left[\frac{(x+1)(x-1)}{(x-1)^2}-\frac{1}{x-1}\right]÷\frac{x^2}{x-1}=\left(\frac{x+1}{x-1}-\frac{1}{x-1}\right)\cdot\frac{x-1}{x^2}=\frac{x+1-1}{x-1}\cdot\frac{x-1}{x^2}=\frac{x}{x-1}\cdot\frac{x-1}{x^2}=\frac{1}{x}$.要使分式有意义,则$x$不能为$1$、$0$.$\because x$的取值范围是$-1\leqslant x\leqslant1$,且为整数,$\therefore x=-1$.当$x=-1$时,原式$=\frac{1}{-1}=-1$.