设a=根号3-3,求[(a+1/a2-a)+(4/1-a2)]除以(a2+2a-3/a2+3a)的值

-√3-2能否写一下过程呢？？？[(a+1)/(a²-a)+4/(1-a²)]/[(a²+2a-3)/(a²+3a)]=[(a+1)/a(a-1)+4/(1-a)(1+a)]/[(a+3)(a-1)/a(a+3)]=[(a+1)/a(a-1)-4/(a-1)(a+1)]/[(a-1)/a]={[(a+1)²/[a(a-1)(a+1)]-4a/[a(a-1)(a+1)]}/[(a-1)/a]=(a-1)²/[a(a-1)(a+1)]/[(a-1)/a]=(a-1)/[a(a+1)]/[(a-1)/a]=1/(a+1)=1/(√3-3+1)=1/(√3-2)=1(√3+2)/(√3-2)(√3+2)=-√3-2