第150页 - 通城学典课时作业本八年级数学苏科版江苏专版

2

$-4$

$ \begin{aligned}解：原式&=\frac {a^2-(a+1)(a-1)}{a-1} \\ &=\frac 1{a-1} \\ \end{aligned}$

$ 解：原式=\frac {(m+2)(2-m)+5}{2-m}×\frac {2m-4}{3-m}$

$ =\frac {9-\ \mathrm {m^2}}{2-m}×\frac {2(m-2)}{3-m}$

$ =-2(3+m)$

$ =-6-2m$

$ \begin{aligned}解：原式&=\frac{x^2-2x+1}{x}·\frac{x}{x^2-1} \\ &=\frac{(x-1)^2}{x}·\frac{x}{(x+1)(x-1)} \\ &=\frac{x-1}{x+1} \\ \end{aligned}$

$ \begin{aligned} 解：2+x(x+1)&=(x+1)(x-1) \\ 2+x^2+x&=x^2-1 \\ x&=-3 \\ \end{aligned}$

$ 检验：当x=-3时，(x+1)(x-1)≠0 $

$ ∴原分式方程的解为x=-3 $

$ \begin{aligned}解：2(a-1)+a+1&=4a \\ a&=-1 \\ \end{aligned}$

$检验：当a=-1时，a(a+1)(a-1)=0$

$∴a=-1是增根，原方程无解$

$解：原式=\frac {a(a+1)+5a+9}{(a+1)(a-1)}×\frac {a-1}{a+3}$

$ =\frac {a^2+6a+9}{(a+1)(a-1)}×\frac {a-1}{a+3}$

$ =\frac {(a+3)^2}{(a+1)(a-1)}×\frac {a-1}{a+3}$

$ =\frac {a+3}{a+1}$