Đáp án+Giải thích các bước giải:
`\frac{4x}{\sqrt{x}+1}`
`=\frac{4x-4+4}{\sqrt{x}+1}`
`=\frac{4(x-1)+4}{\sqrt{x}+1}`
`=\frac{4(\sqrt{x}-1)(\sqrt{x}+1)+4}{\sqrt{x}+1}`
`=4(\sqrt{x}-1)+\frac{4}{\sqrt{x}+1}`
`=4\sqrt{x}-4+\frac{4}{\sqrt{x}+1}`
`=4\sqrt{x}+4+\frac{4}{\sqrt{x}+1}-8`
`=4(\sqrt{x}+1)+\frac{4}{\sqrt{x}+1}-8`
Áp dụng BĐT Cauchy, ta có:
`4(\sqrt{x}+1)+\frac{4}{\sqrt{x}+1}>=2\sqrt{4(\sqrt{x}+1).\frac{4}{\sqrt{x}+1}}=2\sqrt{16}=8`
`->4(\sqrt{x}+1)+\frac{4}{\sqrt{x}+1}-8>=8-8=0`
Dấu "=" xảy ra khi `4(\sqrt{x}+1)=\frac{4}{\sqrt{x}+1}`
`->4(\sqrt{x}+1)^2=4`
`->2(\sqrt{x}+1)=2`
`->\sqrt{x}+1=1`
`->x=0`
Vậy `GTNNNN=0` khi `x=0`
