先化简,再求值:(a2-2ab+b2)/(a2-b2)/(1/a-1/b),其中a=根号2+1,b=根号2-1
问题描述：
先化简,再求值:(a2-2ab+b2)/(a2-b2)/(1/a-1/b),其中a=根号2+1,b=根号2-1
答
(a2-2ab+b2)/(a2-b2)/(1/a-1/b）
=(a-b)²/(a-b)(a+b) /[(b-a)/ab]
=(a-b)/(a+b) *ab/(b-a)
=-ab/(a+b)
=-1/(√2+1+√2-1)
=-1/2√2
=-√2/4
答
:(a2-2ab+b2)/(a2-b2)/(1/a-1/b),
=(a-b)^2÷（a-b)(a+b)÷[(b-a)/(ab)]
=-ab/(a+b)
答
原式=(a-b)²/(a+b)(a-b)÷(b-a)/ab
=(a-b)/(a+b)×[-ab/(a-b)]
=-ab/(a+b)
=-(2-1)/(2√2)
=-√2/4