设实数a、b满足a2-8a+6=0及6b2-8b+1=0,求ab+1/ab的值.

问题描述:

设实数a、b满足a2-8a+6=0及6b2-8b+1=0,求ab+

1
ab
的值.

由于6b2-8b+1=0,则b≠0,则(1b)2−8×1b+6=0,当a≠1b时,则a,1b为方程x2-8x+6=0的两个根,不妨设x1=a,x2=1b,则x1+x2=8,x1x2=6,所以ab+1ab=x1x2+x2x1=(x1+x2)2−2x1x2x1x2=64−126=263,当a=1b时,即ab...