Đáp án:
$Min Q=4\sqrt{2}+6$
Giải thích các bước giải:
$Q=\dfrac{x+2\sqrt{x}}{\sqrt{x}-2}$
$\rightarrow Q=\dfrac{x-4+2\sqrt{x}-4+8}{\sqrt{x}-2}$
$\rightarrow Q=\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)+2(\sqrt{x}-2)+8}{\sqrt{x}-2}$
$\rightarrow Q=\sqrt{x}+2+2+\dfrac{8}{\sqrt{x}-2}$
$\rightarrow Q=\sqrt{x}-2+\dfrac{8}{\sqrt{x}-2}+6$
$\rightarrow Q\ge 2\sqrt{(\sqrt{x}-2)(\dfrac{8}{\sqrt{x}-2})}+6$
$\rightarrow Q\ge 4\sqrt{2}+6$
Dấu = xảy ra khi $\sqrt{x}-2=\dfrac{8}{\sqrt{x}-2}\rightarrow x=2+2\sqrt{2}$
