2 câu trả lời
`#tnvt`
`C=\frac{x+16}{\sqrt{x}+3}(x>=0)`
`=\frac{x-9+25}{\sqrt{x}+3}`
`=\frac{(\sqrt{x}+3)(\sqrt{x}-3)+25}{\sqrt{x}+3}`
`=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}`
`=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6`
Áp dụng BĐT Cauchy cho hai số dương, ta có:
`(\sqrt{x}+3)+\frac{25}{\sqrt{x}+3}>=2\sqrt{(\sqrt{x}+3).\frac{25}{\sqrt{x}+3}}=2\sqrt{25}=10`
`=>\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6>=10-6=4`
Dấu `=` xảy ra khi `\sqrt{x}+3=\frac{25}{\sqrt{x}+3}`
`<=>(\sqrt{x}+3)^2=25`
Mà `\sqrt{x}+3>=3>0`
`=>\sqrt{x}+3=5`
`<=>\sqrt{x}=2`
`<=>x=4(tm)`
Vậy `GTNNNN_C=4` khi `x=4`