第124页 - 启东中学作业本八年级数学人教版

$解:原式=\frac{a^{2}-b^{2}+b^{2}}{a+b}$

$=\frac{a^{2}}{a+b}$

$解:原式=\frac {2x^{2}+x^{2}-4xy+2y^{2}-x^{2}}{x-y}$

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

$=2x-2y$

$解:原式=\frac{3(x-2)+12+9(x+2)}{2(x+2)(x-2)}$

$=\frac{12(x+2)}{2(x+2)(x-2)}$

$=\frac{6}{x-2}$

$解:原式=\frac {x-1+1-(x+1)}{x^{2}-1}$

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

$解:(1)∵b+\frac{1}{b}=-3,∴\frac{3b^{2}-4b+3}{b}=3b-4+\frac{3}{b}=3(b+\frac{1}{b})-4=3×(-3)-4=-13$

$∴\frac{b}{3b^{2}-4b+3}=-\frac{1}{13}$

$(2)∵ x+\frac{1}{x-1}=-5,∴ \frac{x^{2}-3x+3}{x-1}=\frac{x^{2}-1-3(x-1)+1}{x-1}=x+1-3+\frac{1}{x-1}$

$=x+\frac{1}{x-1}-2=-5-2=-7,∴\frac{x-1}{x^{2}-3x+3}=-\frac{1}{7}$

②③

$解:(2)设分式\frac{a}{2a+1}的“美妙分式”为A,则|A-\frac{a}{2a+1}|=3, 所以A=\frac{a}{2a+1}+3或A=\frac{a}{2a+1}-3$

$当A=\frac{a}{2a+1}+3时,A=\frac{a}{2a+1}+3=\frac{a}{2a+1}+\frac{6a+3}{2a+1}=\frac{7a+3}{2a+1}$

$当A=\frac{a}{2a+1}-3时,A=\frac{a}{2a+1}-3=\frac{a}{2a+1}-\frac{6a+3}{2a+1}=\frac{-5a-3}{2a+1}=-\frac{5a+3}{2a+1}$

$故\frac{a}{2a+1}的“美妙分式”为\frac{7a+3}{2a+1}或-\frac{5a+3}{2a+1}$

$(3)（更多请点击查看作业精灵详解）$