Đáp án: $Q\ge -2$
Giải thích các bước giải:
Ta có:
$Q=\sqrt{x}\cdot \dfrac{2(\sqrt{x}-3)}{\sqrt{x}+1}$
$\to Q=\sqrt{x}\cdot \dfrac{2(\sqrt{x}+1-4)}{\sqrt{x}+1}$
$\to Q=\sqrt{x}\cdot \dfrac{2(\sqrt{x}+1)-8}{\sqrt{x}+1}$
$\to Q=\sqrt{x}\cdot (2-\dfrac{8}{\sqrt{x}+1})$
$\to Q=2\sqrt{x}-\dfrac{8\sqrt{x}}{\sqrt{x}+1}$
$\to Q=2(\sqrt{x}+1)-2-\dfrac{8(\sqrt{x}+1)-8}{\sqrt{x}+1}$
$\to Q=2(\sqrt{x}+1)-2-(8-\dfrac{8}{\sqrt{x}+1})$
$\to Q=2(\sqrt{x}+1)-2-8+\dfrac{8}{\sqrt{x}+1}$
$\to Q=2(\sqrt{x}+1)+\dfrac{8}{\sqrt{x}+1}-10$
$\to Q=2((\sqrt{x}+1)+\dfrac{4}{\sqrt{x}+1})-10$
$\to Q\ge 2\cdot 2\sqrt{(\sqrt{x}+1)\cdot\dfrac{4}{\sqrt{x}+1}}-10$
$\to Q\ge 8-10$
$\to Q\ge -2$
Dấu = xảy ra khi $\sqrt{x}+1=\dfrac{4}{\sqrt{x}+1}$
$\to (\sqrt{x}+1)^2=4$
$\to \sqrt{x}+1=2$
$\to\sqrt{x}=1$
$\to x=1$
