Question

若实数a≠b,且a,b满足a2-8a+5=0,b2-8b+5=0,则代数式b-1...

若实数a≠b,且a,b满足a2-8a+5=0,b2-8b+5=0,则代数式b-1a-1+a-1b-1的值为( )A.-20B.2C.2或-20D.2或20

Answer 1

由已知条件可知,a、b为方程x2-8x+5=0的两根,此时△>0,
∴a+b=8,ab=5,
∴
b-1
a-1
+
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
=-20
故选A