第75页 - 苏科版八年级（初二）数学学习与评价答案（上下册）

$解：原式=\frac {a²}{b²}×\frac {25b²}{4a^4}×\frac {a}{5b}\ $

$=\frac {25}{4a²}×\frac {a}{5b}$

$=\frac {5}{4ab}$

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

$=(\frac {1}{a}+\frac {1}{b})÷(\frac {1}{a}-\frac {1}{b})$

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

$解：原式=\frac {9x^4}{16y²}×\frac {2y}{3x}+\frac {x²}{2y²}×\frac {x}{2y²}\ $

$=\frac {3x³}{8y}+\frac {x³}{4y^4}\ $

$=\frac {3x³y³+2x^3}{8y^4}$

$解：原式=\frac {x+2y}{x+y}×\frac {xy}{x+2y}×\frac {xy}{x+y}$

$=\frac {xy}{x+y}×\frac {xy}{x+y}$

$=\frac {x^2y^2}{(x+y)²}$

$解：原式=\frac {2}{3}×\frac {(a+b)²}{(a-b)²}×\frac {a-b}{a+b}-\frac {a²}{(a-b)(a+b)}×\frac {b}{a}$

$=\frac {2}{3}×\frac {a+b}{a-b}-\frac {ab}{(a-b)(a+b)}$

$=\frac {2a²+ab+2b²}{3(a²-b²)}$

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

$ =\frac {2x²+8x}{(x-2)(x+2)}×\frac {(2-x)(2+x)}{x}$

$ =-2x-8$

$②原式=\frac {3x}{x-2}×\frac {4-x²}{x}-\frac {x}{x+2}×\frac {4-x²}{x}$

$ =-3(2+x)-(2-x)$

$ =-2x-8$

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

$解：设甲每小时做x个，则乙每小时做(x-5)个$

$由题意可得：\frac {90}{x}=\frac {60}{x-5}$

$解得x=15$

$所以甲每小时做15个，乙每小时做10个$

$解：原式=\frac {p²+m²+n²}{mnp}$