求函数y=x²-3x+3/x-2﹙x＞2﹚的最小值﹙用均值定理﹚
问题描述：
求函数y=x²-3x+3/x-2﹙x＞2﹚的最小值﹙用均值定理﹚
答
下面有括号?
解y=(x²-3x+3)/(x-2)
=[x(x-2)-x+3]/(x-2)
=[x(x-2)-(x-2)+1]/(x-2)
=x(x-2)/(x-2)-(x-2)/(x-2)+1/(x-2)
=(x-1)+1/(x-2)
=(x-2)+1/(x-2)+1
因为x＞2,x-2＞0
所以(x-2)+1/(x-2)≥2,
即 y=(x-2)+1/(x-2)+1有最小值 2+1=3