8. 先化簡,再求值：$\frac {x^{2}-x}{x^{2}+2x + 1}÷(\frac {2}{x + 1}-\frac {1}{x})$,化簡后,從$-2\lt x\lt3的范圍內(nèi)選擇一個你喜歡的整數(shù)作為x$的值代入求值.
答案：解:原式$=\frac{x(x-1)}{(x+1)^2}÷\frac{2x-(x+1)}{x(x+1)}=\frac{x(x-1)}{(x+1)^2}\cdot\frac{x(x+1)}{x-1}=\frac{x^2}{x+1}$.$\because x\neq-1,x\neq0,x\neq1$,$\therefore$當(dāng)$x=2$時,原式$=\frac{4}{3}$.
9. 已知$\frac {A}{x - 1}-\frac {B}{2 - x}= \frac {2x - 6}{(x - 1)(x - 2)}$,求$A$,$B$的值.
答案：解:$\because\frac{A}{x-1}-\frac{B}{2-x}=\frac{2x-6}{(x-1)(x-2)}$,$\therefore\frac{A}{x-1}+\frac{B}{x-2}=\frac{2x-6}{(x-1)(x-2)}$,$\therefore\frac{A(x-2)+B(x-1)}{(x-1)(x-2)}=\frac{2x-6}{(x-1)(x-2)}$,$\therefore\frac{Ax-2A+Bx-B}{(x-1)(x-2)}=\frac{2x-6}{(x-1)(x-2)}$,$\therefore\frac{(A+B)x-(2A+B)}{(x-1)(x-2)}=\frac{2x-6}{(x-1)(x-2)}$,$\therefore\begin{cases}A+B=2,\\2A+B=6,\end{cases}$解得$\begin{cases}A=4,\\B=-2.\end{cases}$
10. (2024春·揚(yáng)州期末)如果兩個分式$M與N的和為常數(shù)k$,且$k$為正整數(shù),則稱$M與N$互為“和整分式”,常數(shù)$k$稱為“和整值”.如分式$M= \frac {x}{x + 1}$,$N= \frac {1}{x + 1}$,$M + N= \frac {x + 1}{x + 1}= 1$,則$M與N$互為“和整分式”,“和整值”$k = 1$.
(1)已知分式$A= \frac {x - 1}{x - 4}$,$B= \frac {x - 7}{x - 4}$,$A與B$是否互為“和整分式”？若不是,請說明理由；若是,請求出“和整值”$k$.
(2)已知分式$C= \frac {4x - 4}{x - 2}$,$D= \frac {G}{x^{2}-4}$,$C與D$互為“和整分式”,且“和整值”$k = 4$,若$x$為正整數(shù),分式$D$的值也為正整數(shù).
①求$G$所代表的代數(shù)式；
②求$x$的值.
答案：解:
(1)$\because A=\frac{x-1}{x-4},B=\frac{x-7}{x-4}$,$\therefore A+B=\frac{x-1}{x-4}+\frac{x-7}{x-4}=\frac{2x-8}{x-4}=\frac{2(x-4)}{x-4}=2$.$\therefore A$與$B$互為"和整分式","和整值"$k=2$.
(2)①$\because C=\frac{4x-4}{x-2},D=\frac{G}{x^2-4}$,$\therefore C+D=\frac{4x-4}{x-2}+\frac{G}{x^2-4}=\frac{(4x-4)(x+2)}{(x+2)(x-2)}+\frac{G}{(x+2)(x-2)}=\frac{4x^2+4x-8+G}{(x+2)(x-2)}$.$\because C$與$D$互為"和整分式"且"和整值"$k=4$,$\therefore\frac{4x^2+4x-8+G}{(x+2)(x-2)}=4$,即$4x^2+4x-8+G=4(x^2-4)$,$\therefore G=-4x-8$.
②$\because D=\frac{G}{x^2-4}=\frac{-4x-8}{x^2-4}=\frac{-4(x+2)}{(x+2)(x-2)}=-\frac{4}{x-2}$,且若$x$為正整數(shù),分式$D$的值也為正整數(shù),$\therefore x-2=-1$或$x-2=-2$或$x-2=-4$;$\therefore x=1$($x=0$和$x=-2$舍去).
