第84页 - 经纶学典学霸八年级数学上下册【江苏国标】

$\frac{2}{3} $

3

5或10

±2(x+1)或±2(x²-1)

$解:\frac{x²}{x²-9}=\frac{x²(x-3)}{(x+3)(x-3)²}，$

$\frac{3}{6x-9-x²}=\frac{3(x+3)}{(x+3)(x-3)²}.$

$解:\frac{a²-4}{a²-4a+4}= \frac{(a-2)(a+2)}{(a-2)²}=\frac{a+2}{a-2}=\frac{(a+2)²}{(a+2)(a-2)}，$

$\frac{4a}{a²+2a}= \frac{4a}{a(a+2)}= \frac{4}{a+2}=\frac{4(a-2)}{(a+2)(a-2)}.$

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

$\frac{y}{x²+2xy+y²} =\frac{y(x-y)}{(x+y)²(x-y)} ，$

$\frac{2}{y²-x²}= -\frac {2(x+y) }{(x+y)²(x-y)}.$

$解:a-b=\frac{(a-b)²(a+b)}{a²-b²}，$

$\frac{b}{a-b}=\frac{b(a+b)}{a²-b²}，$

$\frac{1}{a²-b²}=\frac{1}{a²-b²}.$

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

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

$由ab=3，a+b=4，得a-b=±2，$

$∴\frac{a-1}{a+1}=\frac{1}{2}或0.$

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

$\frac{x²+3}{x+2}=\frac{x²-4+7}{x+2}=\frac{(x+2)(x-2)+7}{x+2}=x-2+\frac {7}{x+2}.$

$(2)\frac{2x²-1}{x-1}=\frac{2x²-2+1}{x-1}=\frac{2(x+1)(x-1)+1}{x-1}$

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

$∵分式的值为整数，x为整数，$

$∴x-1=1或x-1=-1，解得x=2或x=0，$

$∴当x=2或0时，分式\frac{2x²-1}{x-1}的值为整数.$