第62页 - 苏科版数学补充习题八年级上下册答案

$解：原式 =-{\frac {{b}^{3}} {{a}^{3}{c}^{3}}}÷{\frac {{b}^{4}} {{a}^{4}}}$

$=-{\frac {{b}^{3}} {{a}^{3}{c}^{3}}}·{\frac {{a}^{4}} {{b}^{4}}}$

$=-{\frac {a} {b{c}^{3}}}$

$解：原式 ={\frac {(a+3)(a-3)} {{(a+2)}^{2}}}·{\frac {2(a+2)} {a-3}}·{\frac {a+2} {a+3}}$

$=2$

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

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

$=a$

$解：原式 =\left [ {{\frac {(a+2)(a-2)} {(a-3)(a+2)}+{\frac {a+2} {a-3}}}} \right ]·{\frac {a-3} {a+1}}$

$={\frac {(a-2)+(a+2)} {a-3}}·{\frac {a-3} {a+1}}$

$={\frac {2a} {a-3}}·{\frac {a-3} {a+1}}$

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

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

$={\frac {3{x}^{3}} {4y}}+{\frac {{x}^{3}} {4{y}^{4}}}$

$={\frac {3{x}^{3}{y}^{3}+{x}^{3}} {4{y}^{4}}}$