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