第63页 - 苏科版数学补充习题八年级上下册答案

$解：原式 ={\frac {3-{a}^{2}} {{(a+1)}^{2}}}+{\frac {a+1-1} {a+1}}÷{\frac {a-1+1} {a-1}}$

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

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

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

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

$={\frac {2} {{(a+1)}^{2}}}$

$解：(1)原式 ={\frac {{m}^{2}-9-7} {m+3}}·{\frac {2(m+3)} {m-4}}$

$={\frac {(m+4)(m-4)} {m+3}}·{\frac {2(m+3)} {m-4}}$

$=2m+8$

$当m=-1时，原式=2×(-1)+8=6$

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

$={\frac {2{a}^{2}+4a} {(a+1)(a-1)}}·{\frac {(a+1)(a-1)} {a}}$

$={\frac {a(2a+4)} {(a+1)(a-1)}}·{\frac {(a+1)(a-1)} {a}}$

$=2a+4$

$当a=8时，原式=2×8+4=20$

${x}^{2} -1$

$x=5$

$x=-3$

$~-1$

$~-5$