第72页 - 苏科版数学补充习题八年级上下册答案

C

$解：原式 ={\frac {{a}^{3}{b}^{6}} {-{a}^{5}{b}^{5}}}+{\frac {b} {{a}^{2}}}$

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

$\ \ \ \ \ \ \ \ =0$

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

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

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

$\ \ \ \ \ \ \ \ ={\frac {x+2y} {x+y}}·{\frac {xy} {x+2y}}·{\frac {xy} {x+y}}$

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

$\ \ \ \ \ \ \ \ ={\frac {{x}^{2}{y}^{2}} {{x}^{2}+2xy+{y}^{2}}}$

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

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

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