Question

化简求值:[(a^2-1)/(a^2-2a+1)]+[(2a-a^2)/(a-2)]/a,其中a=根号2+1.

Answer 1

a=根号2+1
[(a^2-1)/(a^2-2a+1)] + [(2a-a^2)/(a-2)] / a
= [(a+1)(a-1)/(a-1)^2] + [2(2-a)/(a-2)] / a
= [(a+1)/(a-1)] + [-2)] / a
= (a+1)/(a-1) - 2 / a
= (根号2+1+1)/(根号2+1-1) - 2 / (根号2+1）
= (根号2+2)/根号2 - 2(根号2-1) / (2-1）
= (1+根号2) - 2根号2 +2
= 3 - 根号2

追问

如果不代入a，式子是怎样的

追答

我试过了，通过那一步代入a是最简单的，如果不代入a继续通分反而繁琐。
什么事都要讲究效率是吧！

追问

题目说先化简

Answer 2

解答：
:[(a^2-1)/(a^2-2a+1)]+[(2a-a^2)/(a-2)]/a
=(a+1)(a-1)/(a-1)^2+2a(1-a)/[a(a-2)]
=(a+1)/(a-1)-(a-1)/(a-2)
=[(a+1)(a-2)-(a-1)^2]/[(a-1)(a-2)]
=[a^2-a-2-a^2+4a-4]/[(a-1)(a-2)]
=(3a-6)/[(a-1)(a-2)]=3/(a-1)=3/[根号2]=(3倍根号2)/2

Answer 3

[(a^2-1)/(a^2-2a+1)]+[(2a-a^2)/(a-2)]/a=[(a-1)(a+1)/(a-1)^2]+[a(2-a)/(a-2)]/a
=[(a+1)/(a-1)]+(-a)/a=[(a+1)/(a-1)]-1=[(√2+1+1)/(√2+1-1)]-1=(√2+2)/√2-1=√2