解：设方程x^2-7x+8=0的两根为x1、x2，则x1+x2=7x1x2=8,即-1/x^2-1/x2^2=-[(x1+x2)^2-2x1x2]/(x1x2)^2=-(49-16)/64=-33/64(-1/x^2)(-1/x2^2)=[1/(x1x2)]^2=1/64,所以方程为x^2+33x/64+1/64=0,即64x^2+33x+1=0即为所求是否可以解决您的问题？
1/x-7/根号x+8=0
