题目

a不等于b,且a,b满足a^2-8a+5=0,b^2-8b+5=0,则[（b-1)/(a-1)+(a-1)/(b-1)]的值是多少

提问时间：2020-12-06

答案

a^2-8a+5=0,b^2-8b+5=0,且a不等于b
所以a和b是方程x^2-8x+5=0的两个跟
所以由韦达定理
ab=5,a+b=8
(b-1)/(a-1)+(a-1)/(b-1)
=[(b-1)^2+(a-1)^2]/(a-1)(b-1)
=(a^2-2a+1+b^2-2b+1)/(ab-a-b+1)
=(a^2+b^2-2a-2b+2)/(ab-a-b+1)
=[(a+b)^2-2ab-2(a+b)+2]/[ab-(a+b)+1]
=(8^2-2*5-2*8+2)/(5-8+1)
=40/(-2)
=-20