Đáp án+Giải thích các bước giải:
`\frac{3x}{\sqrt{x}+1}(x>=0)`
`=\frac{3x+3\sqrt{x}-3\sqrt{x}-3+3}{\sqrt{x}+1}`
`=\frac{3\sqrt{x}(\sqrt{x}+1)-3(\sqrt{x}+1)+3]{\sqrt{x}+1}`
`=3\sqrt{x}-3+\frac{3}{\sqrt{x}+1}`
`=3\sqrt{x}+3+\frac{3}{\sqrt{x}+1}-6`
`=3(\sqrt{x}+1)+\frac{3}{\sqrt{x}+1}-6`
Áp dụng BĐT Cauchy cho hai số dương, ta có:
`3(\sqrt{x}+1)+\frac{3}{\sqrt{x}+1}>=2\sqrt{3(\sqrt{x}+1).\frac{3}{\sqrt{x}+1}}=2\sqrt{9}=6`
`=>3(\sqrt{x}+1)+\frac{3}{\sqrt{x}+1}-6>=6-6=0`
Dấu `=` xảy ra khi `3(\sqrt{x}+1)=\frac{3}{\sqrt{x}+1}`
`<=>3(\sqrt{x}+1)^2=3`
`<=>(\sqrt{x}+1)^2=1`
`<=>\sqrt{x}+1=1`
`<=>\sqrt{x}=0`
`<=>x=0`
Vậy `GTNNNN=0` khi `x=0`
