Đáp án: $\left(\sqrt{x}-\dfrac32-\dfrac{\sqrt{17}}{2}\right)\left(\sqrt{x}-\dfrac32+\dfrac{\sqrt{17}}{2}\right)$
Giải thích các bước giải:
ĐKXĐ: $x\ge 0$
Ta có:
$x-3\sqrt{x}-2$
$=\left(x-2\cdot\sqrt{x}\cdot\dfrac32+\left(\dfrac32\right)^2\right)-\left(\dfrac32\right)^2-2$
$=\left(\sqrt{x}-\dfrac32\right)^2-\dfrac{17}{4}$
$=\left(\sqrt{x}-\dfrac32\right)^2-\left(\dfrac{\sqrt{17}}{2}\right)^2$
$=\left(\sqrt{x}-\dfrac32-\dfrac{\sqrt{17}}{2}\right)\left(\sqrt{x}-\dfrac32+\dfrac{\sqrt{17}}{2}\right)$
