Question

若x>1 求函数y=x^2/x-1的最小值 。详解

Answer 1

y=x^2/(x-1)
=(x^2-1)/(x-1)+1/(x-1)
=x+1+1/(x-1)
=x-1+1/(x-1)+2（磨掘用均值不等式）
大于等于瞎厅核4（x=2时取等号）
若x>伏宽1
求函数y=x^2/x-1的最小值为4

Answer 2

y=2x+[1/薯绝(x-1)]
=2(x-1)+[1/(x-1)]+2
>=2+2(根号2)
y的最小值=2+2(根号纳锋2)，y取最小值时，2(x-1)=1/(x-1),
x=1+(1/2)(根号数茄姿2)

Answer 3

将原式变化：
y=x²/(x-1)=((x²-1)+1)/(x-1)=((x+1)(x-1)+1)/(x-1)
因为x>1,x-1≠0
y=(x+1)+1/(x-1)=(x-1)+1/(x-1)+2
x-1>0
根据不等李友式性质：
(x-1)+1/(x-1)>2√(x-1)*1/(x-1)
(x-1)+1/(x-1)>2
y>哪昌槐4
所以最小值为迅散4。

Answer 4

即x-1>0
所以芹晌y=(x²-1+1)/(x-1)
=(x²-1)/(x-1)+1/(x-1)
=x+1+1/(x-1)
=(x-1)+1/腊首蚂(x-1)+2≥2√[(x-1)*1/(x-1)]+2=4
所以轮埋最小值是4