電子課本網(wǎng) › 第118頁(yè)

第118頁(yè)

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0

$\frac{3x}{x-4}$

$\frac {x+2y}{x-2y}$

$\frac{2}{5}$

$解:\frac{1}{3x^{2}}=\frac {4y}{12x^{2}y}$

$\frac{5}{12xy}=\frac{5x}{12x^{2}y}\ $

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

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

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

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

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

$解:原式=\frac {x}{3x-y}$

$當(dāng)x=\frac{4}{3},y=-\frac{2}{3}時(shí)，原式=\frac{2}{7}$

C

$解:\frac{m^{2}+3}{m+1}=\frac{m^{2}-1+4}{m+1}=\frac{m^{2}-1}{m+1}+\frac{4}{m+1}=m-1+\frac{4}{m+1}$

5

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

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

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

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