第59页 - 苏科版数学补充习题八年级上下册答案

不正确

分母不能去掉

${\frac {4x} {{x}^{2}-1}}$

$解：原式=\frac{a-1}{a-1}$

$=1$

$解：原式 ={\frac {4b} {6a{b}^{2}}}+{\frac {a} {6a{b}^{2}}}$

$\ \ \ \ \ \ \ \ ={\frac {4b+a} {6a{b}^{2}}}$

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

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

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

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

$解：原式 ={\frac {(a+4)(a+2)} {{(a+2)}^{2}}}+{\frac {4a-{a}^{2}} {{(a+2)}^{2}}}$

$\ \ \ \ \ \ \ \ ={\frac {({a}^{2}+6a+8)+(4a-{a}^{2})} {{(a+2)}^{2}}}$

$\ \ \ \ \ \ \ \ ={\frac {10a+8} {{a}^{2}+4a+4}}$

$解：原式 ={\frac {2a-3} {(2a+3)(2a-3)}}-{\frac {3(2a+3)} {(2a+3)(2a-3)}}-{\frac {2a+15} {(2a+3)(2a-3)}}$

$ ={\frac {(2a-3)-3(2a+3)-(2a+15)} {(2a+3)(2a-3)}}$

$ ={\frac {-6a-27} {{4a}^{2}-9}}$

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

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

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

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