第85页 - 经纶学典新课时作业八年级数学上下册【江苏国标】

C

$-\frac{1}{m+1}$

$\frac{x^2}{1-x}$

$\frac{4b}{a²+4a}$

$ \begin{aligned}解:原式&=\frac{3(x-2)}{(x-2)²}+\frac{5-x²}{(x-2)²}+\frac{(x-1)(x-2)}{(x-2)²} \\ &=\frac{1}{(x-2)²}， \\ \end{aligned}$

$将 x=4代入，∴原式=\frac{1}{4}$

（更多请点击查看作业精灵详解）

$解:∵\frac{A}{x-1}+\frac{B}{x+3}=\frac{A(x+3)+B(x-1)}{(x-1)(x+3)}=\frac{(A+B)x+3A-B}{(x-1)(x+3)}，$

$∴\frac {(A+B)x+3A-B}{(x-1)(x+3)}=\frac {x-5}{(x-1)(x+3)}，$

$∴\begin{cases}{A+B=1,}\\{3A-B=-5,}\end{cases}解得\begin{cases}{A=-1,}\\{B=2.}\end{cases}$

$解:(1)\frac{2x-3}{x-3}=\frac{2(x-3)+3}{x-3}=2+\frac {3}{x-3}$

$(2)\frac{x²}{x+1}=\frac{x²-1+1}{x+1}=\frac{(x-1)(x+1)+1}{x+1}=x-1+\frac {1}{x+1}\ $

$∵分式的值为整数，且x为整数，\ $

$∴x+1=±1，∴x=-2或0.$

$ \begin{aligned}解:原式&=\frac {2n(m-2n)}{m²-4n²}-\frac {m(m+2n)}{m²-4n²}-\frac {4mn}{m²-4n²} \\ &=\frac {2mn-4n²-m²-2mn-4mn}{m²-4n²} \\ &=\frac {-m²-4mn-4n²}{m²-4n²} \\ &=\frac {-(m+2n)²}{(m+2n)(m-2n)} \\ &=-\frac {m+2n}{m-2n} \\ \end{aligned}$

$∵m=3n，$

$ \begin{aligned}∴原式&=-\frac{m+2n}{m-2n} \\ &=-\frac{3n+2n}{3n-2n} \\ &=-5. \\ \end{aligned}$