Question

若实数a不等于b,且a,b满足a2-8a+5=0,b2-8b+5=0,则代数式（b-1)/(a-1)+(a-1)/(b-1)

Answer 1

若实数a不等于b,且a,b满足a2-8a+5=0,b2-8b+5=0 则a+b=8,ab=5
（b-1)/(a-1)+(a-1)/(b-1)=(b-1)2+(a-1)2/(a-1)(b-1)=b2-2b+1+a2-2a+1/ab-a-b+1
=（a+b)2-2ab-2(a+b)+2/ab-(a+b)+1
代入a+b=8,ab=5
则得答案 -20

Answer 2

实数a不等于b,且a,b满足a2-8a+5=0,b2-8b+5=0,
则a,b分别是方程x²-8x+5=0的两个实数根
a+b=8
ab=5
（b-1)/(a-1)+(a-1)/(b-1)
=[(b-1)²+(a-1)²]/(a-1)(b-1)
=[a²+b²-2(a+b)+2]/[ab-(a+b)+1]
=[(a+b)²-2ab-2(a+b)+2]/[ab-(a+b)+1]
=(8²-2*8-2*5+2)/(5-8+1)
=-20