ĐK: $x\ge 0; 2x+1+\sqrt{x}\ge 0$
$2x+1+\sqrt{x}\ge 0$
$\to 2.(\sqrt{x})^2+\sqrt{x}+1\ge 0$
$\to (\sqrt{x})^2+\dfrac{1}{2}\sqrt{x}+\dfrac{1}{2}\ge 0$
$\to (\sqrt{x})^2+2\sqrt{x}.\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{7}{16}\ge 0$
$\to \Big(\sqrt{x}+\dfrac{1}{4}\Big)^2+\dfrac{7}{16}\ge 0$ (luôn đúng)
Vậy ĐKXĐ: $x\ge 0$
