Question

x＞1，求y＝x²－x＋2/x－1最小值

Answer 1

x>1
所以x-1>0
x+2/(x-1)=(x-1)+2/(x-1)+1>=2倍根号下[(x-1)*2/(x-1)]+1=2倍根号下2+1
当且仅当x-1=2/x-1即x=根号2+1时等号成立
所以y=x+2/(x-1)的最小值为2倍根号下2+1

Answer 2

y=(x²-x+2)/(x-1)
=(x-1)
+2/(x-1)
+1
x>1，x-1>0
由
基本不等式
得：(x-1)+2/(x-1)≥2√[(x-1)·2/(x-1)]=2√2
(x-1)
+2/(x-1)
+1≥1+2√2
y≥1+2√2
y的最小值为1+2√2