第105页 - 通城学典课时作业本八年级数学人教版南通专版

A

0

$\frac {x-y}{xy}$

$-\frac {17}{4}$

$解：原式=\frac {{a}^2-4}{a-2}-\frac {{a}^2+4}{a-2}$

$=\frac {{a}^2-4-({a}^2+4)}{a-2}$

$=-\frac 8{a-2}$

$解：原式=\frac {x+1}{{x}^2-1}+\frac {{x}^2-3x}{{x}^2-1}$

$=\frac {x+1+{x}^2-3x}{{x}^2-1}$

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

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

$=\frac {x-1}{x+1}$

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

$=\frac 1{x-y}+\frac 1 {x-y}+\frac 2 {y-x}$

$=\frac 1{x-y}+\frac 1 {x-y}-\frac 2 {x-y}$

$=0$

解：根据题意，得$\frac A{x-1}-\frac B{2-x}=\frac {A(x-2)+B(x-1)}{(x-1)(x-2)}=\frac {(A+B)x-2A-B}{(x-1)(x-2)}=\frac {2x-6}{(x-1)(x-2)}$

$∴{{\begin{cases} {{A+B=2}} \\{-2A-B=-6} \end{cases}}}$

解得，${{\begin{cases} {{A=4}} \\{B=-2} \end{cases}}}$

解：由题意可得$\frac {M}{a+1}=\frac {a^2}{a(a+1)}=\frac {a}{a+1}，$

则$M=a，$

那么$\frac {a}{a+1}-\frac {1}{a^2+a}$

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

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

$=\frac {(a+1)(a-1)}{a(a+1)}$

$=\frac {a-1}{a}，$

当$a=100$时，

原式$=\frac {100-1}{100}=\frac {99}{100}.$