第75页 - 苏科版八年级数学课课练答案（上下册）

$解：原式=4x²y÷\frac {4x²}{y²}$

$ =4x²y×\frac {y²}{4x²}$

$ =y³$

$解：原式=\frac {3(a-b)}{10ab}×\frac {15a²b²}{(a+b)(a-b)}$

$ =\frac {9ab}{2(a+b)}$

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

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

$解：原式=\frac {2(x-3)}{(x-2)²}×\frac {1}{x+3}×\frac {(x+3)(x-2)}{3-x}$

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

$解：原式=\frac {x+3y}{(x+2y)(x-2y)}×\frac {x+2y}{(x+3y)²}$

$ =\frac {1}{(x-2y)(x+3y)}$

$ 因为x²+y²-4x+6y+13=0$

$ 所以(x-2)²+(y+3)²=0$

$ 所以x=2，y=-3$

$ 所以原式=\frac {1}{(2+6)×(2-9)}=-\frac {1}{56}$

$解：原式=[\frac {x+2}{x(x-2)}-\frac {x-1}{(x-2)²}]×\frac {x}{x-4}$

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

$ =\frac {1}{(x-2)²}$