(1+x)/(x^2+x-2)=(1+x)/[(x-1)*(x+2)]
(x^2-1)/(x+2)=[(x-1)*(x+1)]/(x+2)
所以原式={(1+x)/[(x-1)*(x+2)]}/{[(x-1)*(x+1)]/(x+2)}
={(1+x)/[(x-1)*(x+2)]}*{(x+2)/[(x-1)*(x+1)]}
=[(1+x)*(1+2)]/{(x-1)^2*(x+1)*(x+2)}
=1/(x-1)^2
将x=1/2代入
原式=4
(1+x)/(x^2+x-2)=(1+x)/[(x-1)*(x+2)]
(x^2-1)/(x+2)=[(x-1)*(x+1)]/(x+2)
所以原式={(1+x)/[(x-1)*(x+2)]}/{[(x-1)*(x+1)]/(x+2)}
={(1+x)/[(x-1)*(x+2)]}*{(x+2)/[(x-1)*(x+1)]}
=[(1+x)*(1+2)]/{(x-1)^2*(x+1)*(x+2)}
=1/(x-1)^2
将x=1/2代入
原式=4