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$\frac{2}{3} $

3

5或10

±2(x+1)或±2(x2-1)

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

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

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

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

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

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

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

$解:a-b=\frac{(a-b)2(a+b)}{a2-b2}，$

$\frac{a-b}=\frac{b(a+b)}{a2-b2}，$

$\frac{1}{a2-b2}=\frac{1}{a2-b2}.$

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

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

$由ab=3，a+b=4，得a-b=±2，$

$∴\frac{a-1}{a+1}=\frac{1}{2}或0.$

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

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

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

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

$∵分式的值為整數(shù)，x為整數(shù)，$

$∴x-1=1或x-1=-1，解得x=2或x=0，$

$∴當x=2或0時，分式\frac{2x2-1}{x-1}的值為整數(shù).$

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