双曲线y=5/x与直线y=x-7有一个交点为(a,b),
所以
b=5/a (1)
b=a-7 (2)
ab=5
a-b=7
a/b+b/a
=(a^2+b^2)/ab
=[(a-b)^2+2ab]/ab
=(a-b)^2/ab +2
=49/5+2
=59/5
双曲线y=5/x与直线y=x-7有一个交点为(a,b),
所以
b=5/a (1)
b=a-7 (2)
ab=5
a-b=7
a/b+b/a
=(a^2+b^2)/ab
=[(a-b)^2+2ab]/ab
=(a-b)^2/ab +2
=49/5+2
=59/5