第84页 - 经纶学典学霸八年级数学上下册【江苏国标】

$±2(x+1)或±2(x^{2}-1)$

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

$\frac{3}{6x-9-x^{2}}=-\frac{3(x+3)}{(x+3)(x-3)^{2}}$

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

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

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

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

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

$解:a-b=\frac{(a-b)^{2}(a+b)}{a^{2}-b^{2}}$

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

$\frac{1}{a^{2}-b^{2}}=\frac{1}{a^{2}-b^{2}}$

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

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

$(2)由(1)得\frac{a-1}{a+1}=\frac{(a-1)(b+1)}{(a+1)(b+1)}=\frac{ab+a-b-1}{ab+a+b+1}$

$由ab=3，a+b=4,得a-b=±2$

$∴\frac{a-1}{a+1}=\frac{1}{2}或0$

$解:(1)\frac{3x+1}{x-1}=\frac{3x-3+4}{x-1}=3+\frac{4}{x-1}$

$\frac{x^{2}+3}{x+2}=\frac{(x+2)(x-2)+7}{x+2}$

$=x-2+\frac {7}{x+2}$

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

$∵分式的值为整数，x为整数$

$∴x-1=1或x-1=-1，解得x=2或x=0$

$∴当x=2或0时，分式\frac{2x^{2}-1}{x-1}的值为整数$