Question

tìm min \(\dfrac{x+16}{\sqrt{x}+3}\)

Answer 1

ĐKXĐ: `x>=0`

\(\dfrac{x+16}{\sqrt{x}+3}=\dfrac{\left(x-9\right)+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}\\ =\left(\sqrt{x}+3\right)+\dfrac{25}{\sqrt{x}+3}-6\)

Với \(x\ge0=>\left\{{}\begin{matrix}\sqrt{x}+3>0\\\dfrac{1}{\sqrt{x}+3}>0\end{matrix}\right.\)

Áp dụng bđt cô-si ta có:

\(\left(\sqrt{x}+3\right)+\dfrac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right)\cdot\dfrac{25}{\sqrt{x}+3}}-6=2\cdot5-6=4\)

Dấu "=" xảy ra: \(\sqrt{x}+3=\dfrac{25}{\sqrt{x}+3}\Leftrightarrow\left(\sqrt{x}+3\right)^2=25\Leftrightarrow x=4\left(tm\right)\)