第70页 - 江苏版实验班提优训练数学八年级上下册答案

A

B

x

$ \begin{aligned}解:原式&=\frac {2x-3-x+1}{x-2} \\ &=\frac {x-2}{x-2} \\ &=1 \\ \end{aligned}$

$ \begin{aligned} 解:原式&=\frac {(x+3y)-(x+2y)+(2x-3y)}{x^{2} -y^{2} } \\ &=\frac {x+3y-x-2y+2x-3y}{x^{2} -y^{2} } \\ &=\frac {2(x-y)}{(x+y)(x-y)} \\ &=\frac{2}{x+y} \\ \end{aligned}$

$ \begin{aligned} 解:原式&=\frac {m(m+n)}{(m-n)(m+n)}-\frac {n(m-n)}{(m-n)(m+n)}+\frac {2mn}{(m-n)(m+n)} \\ &=\frac {(m+n)^{2} }{(m-n)(m+n)} \\ &=\frac {m+n}{m-n} \\ \end{aligned}$

$ \begin{aligned} 解:原式&=\frac {x^{2} }{x+y}-\frac {(x+y)(x-y)}{x+y} \\ &=\frac {x^{2} -x^{2} +y^{2} }{x+y} \\ &=\frac {y^{2} }{x+y} \\ \end{aligned}$

$ \begin{aligned}解:原式&=\frac {3x(2y-3z)+2y(2z-3x)+z(9x-4y)}{6xyz} \\ &=\frac {6xy-9xz+4yz-6xy+9zx-4yz}{6xyz} \\ &=0 \\ \end{aligned}$

$ \begin{aligned}解:原式&=1-\frac {4x(2x-y)+(2x+y)^{2} }{4x^{2} -y^{2} } \\ &=-\frac {2(2x-y)^{2} }{(2x+y)(2x-y)} \\ &=-\frac{2(2x+y)}{2x-y} \\ \end{aligned}$

B

A