函数y=(x²+8)/(x-1) (x＞1)的最小值是函数y=(x²+8)/(x-1) (x＞1)的最小值是
问题描述：
函数y=(x²+8)/(x-1) (x＞1)的最小值是
函数y=(x²+8)/(x-1) (x＞1)的最小值是
答
解
y=[x^2+8]/(x-1)
=[（x-1）^2+2(x-1)+9]/(x-1)
=（x-1）+9/(x-1)+2
≥2√(x-1)×9/(x-1)+2
=6+2
=8
故函数y=(x²+8)/(x-1) (x＞1)的最小值是8.
答
x>1即x-1>0
y=(x²-1)/(x-1)+9/(x-1)
=x+1+9/(x-1)
=(x-1)+9/(x-1)+2≥2√[(x-1)*9/(x-1)]+2=8
所以最小值是8