第76页 - 苏科版八年级数学课课练答案（上下册）

$解：因为\frac {1}{x}-\frac {1}{y}=3$

$ \frac {1}{x}-\frac {1}{y}=\frac {y-x}{xy}=3$

$ 所以y-x=3xy$

$ 原式=\frac {x-(y-x)-y}{2x+y-x-2y}$

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

$ =2$

D

A

$解：原式=\frac {a-b}{a-b}×\frac {ab}{a-b}$

$ =\frac {ab}{a-b}$

$解：原式=\frac {x(x+2)-x(x-2)}{(x-2)(x+2)}×\frac {2-x}{4x}$

$ =\frac {4x}{(x+2)(x-2)}×\frac {2-x}{4x}$

$ =-\frac {1}{x+2}$

$解：原式=1-\frac {a-b}{a+2b}×\frac {(a+2b)²}{(a-b)(a+b)}$

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

$ =-\frac {b}{a+b}$

$解：原式=\frac {a²-b²+2b²}{a-b}$

$ =\frac {a²+b²}{a-b}$

$解：原式=\frac {(x-y)²+4xy}{x-y}×\frac {(x+y)²-4xy}{x+y}$

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

$=(x+y)(x-y)$

$ =x²-y²$

$解：原式=\frac {m+2-(m-2)}{(m+2)(m-2)}+\frac {2(m-1)-2(m+1)}{(m+1)(m-1)}$

$ =\frac {4}{(m+2)(m-2)}-\frac {4}{(m+1)(m-1)}$

$ =\frac {12}{(m+2)(m-2)(m+1)(m-1)}$