一道数学问题

一道初二数学题，急用！先化简，再

先化简，再求值：(1+x)/（x^２＋x-2)除以（x^2-1)/(x+2),其中x=1/2。 解:：[(1+x)/（x^２＋x-2)]÷[（x^2-1)/(x+2)] =[(1+x)/（x+2)(x-1)]÷[（x-1)(x+1)/(x+2)] =[(1+x)/（x+2)(x-1)]×[(x+2)/（x-1)(x+1)] =[(1+x)(x+2)]/[（x+2)(x-1)（x-1)(x+1)] =1/[(x-1)(x-1)](x=1/2) =1/[(-1/2)(-1/2)] =1/(1/4) =4 。

  先化简，再求值：(1+x)/（x^２＋x-2)除以（x^2-1)/(x+2),其中x=1/2。
   解:：[(1+x)/（x^２＋x-2)]÷[（x^2-1)/(x+2)] =[(1+x)/（x+2)(x-1)]÷[（x-1)(x+1)/(x+2)] =[(1+x)/（x+2)(x-1)]×[(x+2)/（x-1)(x+1)] =[(1+x)(x+2)]/[（x+2)(x-1)（x-1)(x+1)] =1/[(x-1)(x-1)](x=1/2) =1/[(-1/2)(-1/2)] =1/(1/4) =4 。收起