若实数a、b满足a2-8a+5=0，b2-8b+5=0，则b−1a−1+a−1b−1的值为．

Question

若实数a、b满足a²-8a+5=0，b²-8b+5=0，则(b−1)/(a−1)+(a−1)/(b−1)的值为______．

Answer 1

当a≠b时，由实数a、b满足a²-8a+5=0，b²-8b+5=0，可把a，b看成是方程x²-8x+5=0的两个根，
∴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)
=(64−10−16+2)/(5−8+1)=(40)/(−2)=-20，
当a=b≠1时，∴(b−1)/(a−1)+(a−1)/(b−1)=(a−1)/(a−1)+(a−1)/(a−1)=1+1=2，
故答案为：-20或2．