一道初二化简的题!

题目

一道初二化简的题!
（x²-x-2分之3x+2）+（1-（x+1分之1））÷（1+（x-1分之1））

答案

(3x+2)/(x²-x-2)+[1-1/(x+1)]÷[1+(1/(x-1)]
=(3x+2)/(x-2)(x+1)+[(x+1-1)/(x+1)]÷[(x-1+1)/(x-1)]
=(3x+2)/(x-2)(x+1)+[x/(x+1)]÷[x/(x-1)]
=(3x+2)/(x-2)(x+1)+x/(x+1)*(x-1)/x
=(3x+2)/(x-2)(x+1)+(x-1)/(x+1)
=(3x+2)/(x-2)(x+1)+(x-1)(x-2)/(x-2)(x+1)
=(3x+2)/(x-2)(x+1)+(x^2-3x+2)/(x-2)(x+1)
=(3x+2+x^2-3x+2)/(x-2)(x+1)
=(x^2+4)/(x-2)(x+1)