函数y=(x^2+2)/(x-1)(x>1)的最小值对函数做相应变化y=(x^2+2)/(x-1)y=(x^2-1+3)/(x-1)y=[(x-1)(x+1)+3]/(x-1)y=x+1+3/(x-1)y=(x-1)+3/(x-1)+2有均值不等式x-1+3/(x-1)>=2√3当且仅当x-1=3/(x-1),即x=√3+1(满足题设x>1)时
函数y=(x^2+2)/(x-1)(x>1)的最小值
对函数做相应变化
y=(x^2+2)/(x-1)
y=(x^2-1+3)/(x-1)
y=[(x-1)(x+1)+3]/(x-1)
y=x+1+3/(x-1)
y=(x-1)+3/(x-1)+2
有均值不等式x-1+3/(x-1)>=2√3当且仅当x-1=3/(x-1),即x=√3+1(满足题设x>1)时等式成立.
所以原函数最小值为2√3+2.
这是他们的回答有两步我看不懂、
y=[(x-1)(x+1)+3]/(x-1)
y=x+1+3/(x-1)
y=(x-1)+3/(x-1)+2
这3步怎么换出来的?、/