Đáp án:
$\begin{array}{l}
1)x \ge 0\\
a)x - 16 = {\left( {\sqrt x } \right)^2} - 16 = \left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)\\
b)x\sqrt x + 8\\
= {\left( {\sqrt x } \right)^3} + {2^3}\\
= \left( {\sqrt x + 2} \right)\left( {x - 2\sqrt x + 4} \right)\\
c)x - 4 = \left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)\\
d)x\sqrt x - 1\\
= {\left( {\sqrt x } \right)^3} - 1\\
= \left( {\sqrt x - 1} \right)\left( {x + \sqrt x + 1} \right)\\
B2)x \ge 0\\
a)x + 2\sqrt x = \sqrt x \left( {\sqrt x + 2} \right)\\
b)x - 3\sqrt x = \sqrt x \left( {\sqrt x - 3} \right)\\
c)x - \sqrt x - 2\\
= x - 2\sqrt x + \sqrt x - 2\\
= \left( {\sqrt x - 2} \right)\left( {\sqrt x + 1} \right)\\
d)x - 4\sqrt x + 3\\
= x - 3\sqrt x - \sqrt x + 3\\
= \left( {\sqrt x - 3} \right)\left( {\sqrt x - 1} \right)
\end{array}$
