第83页 - 通城学典课时作业本八年级数学苏科版江苏专版

$ \begin{aligned}解：原式&=\frac {2(m-3)}{(m+3)(m-3)}×\frac {m+3}{2(m+1)}-\frac m{m+1} \\ &=\frac 1{m+1}-\frac {m}{m+1} \\ &=\frac {1-m}{m+1} \\ \end{aligned}$

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

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

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

$=\frac 1{x-1}$

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

$ \begin{aligned}解：原式&=\frac {3a-(a+1)}{(a+1)(a-1)}÷\frac {2a-1}{a+1} \\ &=\frac {2a-1}{(a+1)(a-1)}×\frac {a+1}{2a-1} \\ &=\frac 1{a-1} \\ \end{aligned}$

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

$=\frac{x+2}{x}$

$当x=-3时，原式=\frac{1}{3}$

$解：(1)f(\mathrm {x})+f(\frac 1{x})=\frac {2x}{x+1}+\frac {2×\frac 1{x}}{\frac 1{x}+1}=\frac {2x}{x+1}+\frac 2{1+x}=\frac {2x+2}{x+1}=\frac {2(x+1)}{x+1}=2$

$(2)由(1)，得f(2)+f(\frac 12)=2，f(3)+f(\frac 13)=2，f(4)+f(\frac 14)=2，···，f(101)+f(\frac 1{101})=2$

$∴原式=2×100+f(1)=200+\frac {2×1}{1+1}=201$