求Y=x-1/x^2+8(X>1)的最小值

Question

求Y=x-1/x^2+8(X>1)的最小值

Answer 1

y=x^2/(x-1)+8 =(x^2-1)/(x-1)+1/(x-1)+8 =(x+1)+1/(x-1)+8 =(x-1)+1/(x-1)+10 因为x>1,所以上式≥2+10=12 所以最小值是12