若有x＞1,求函数y=（2x²-x+1）/（x-1）的最小值

(2x+1)(x-1) +2 2 2
y= ----------------------- = 2x+1 + -------------- = 2(x-1) + -------------- +1
x-1 x-1 x-1
2
≥ 2*√(2(x-1)*--------) +1 =5
x-1
当且仅当
2
2(x-1) = ---------- 时等号成立
x-1
即 x=2时,y有最小值5 再答： 不客气