$\left ( \sqrt{x} - \frac{1}{2} \right )\left ( \sqrt{x} + 2 \right ) = 0$ (ĐK: $x \geq 0$)
$\Leftrightarrow \sqrt{x} - \frac{1}{2} = 0$ (vì $\sqrt{x} + 2 > 0$ với mọi $x$)
$\Leftrightarrow \sqrt{x} = \frac{1}{2}$
$\Leftrightarrow x = \frac{1}{4} (TM)$