第63页 - 苏科版数学补充习题八年级上下册答案

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

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

$\ \ \ \ \ \ \ \ \ \ \ =2m+8$

$当m=-1时，原式=2×(-1)+8=6$

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

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

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

$\ \ \ \ \ \ \ =2a+4$

$当a=8时，原式=2×8+4=20$

${x}^{2} -1$

$x=5$

$x=-3$

$~-1$

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