Question

a平方-2a+1分之a平方-1+a-2分之2a-a平方 其中a=根号2+1

Answer 1

a平方-2a+1分之a平方-1+a-2分之2a-a平方 其中a=根号2+1
=(a^2-1)/(a^2-2a+1)+(2a-a^2)/(a-2)
=(a+1)(a-1)/(a-1)^2-a(a-2)/(a-2)
=(a+1)/(a-1)-a
=(a+1-a(a-1))/(a-1)
=(a+1-a^2+a)/(a-1)
=-(a^2-2a-1)/(a-1)
=-((√2+1)^2-2(√2+1)-1)/((√2+1)-1)
=-(2+2√2+1-2√2-2-1)/√2
=0

Answer 2

解：（a^2-1)/(a^2-2a+1)-(2a-a^2)/(a-2)
=(a+1)(a-1)/(a-1)^2-a(2-a)/(a-2)
=(a+1)/(a-1)+a
=(a^2+1)/(a-1)
={[2^(1/2)+1]^2+1}/[2^(1/2)]
=2^(3/2)+2

追问

最后答案是2×3分之2+2么？

追答

不是
是2的二分之三次方+2
或者是2×根号2+2

抱歉，式子看错了，你参考下面的求解过程吧。
（a^2-1)/(a^2-2a+1)+(2a-a^2)/(a-2)
=(a+1)(a-1)/(a-1)^2+a(2-a)/(a-2)
=(a+1)/(a-1)-a
=(-a^2+2a+1)/(a-1)
=-(a^2-2a+1-2)/(a-1)
=-(a-1)+2/(a-1)
=-√2+√2
=0