Question

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

急啊~~·

Answer 1

-√3-2

追问

能否写一下过程呢？？？

追答

[(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