Đáp án:
$\begin{array}{l}
a)x\sqrt x + \sqrt x - x - 1\\
= \sqrt x \left( {x + 1} \right) - \left( {x + 1} \right)\\
= \left( {x + 1} \right)\left( {\sqrt x - 1} \right)\\
b)\sqrt {ab} + 2\sqrt a + 3\sqrt b + 6\\
= \sqrt a \left( {\sqrt b + 2} \right) + 3\left( {\sqrt b + 2} \right)\\
= \left( {\sqrt b + 2} \right)\left( {\sqrt a + 3} \right)\\
c)3\sqrt a - 2a - 1\\
= 2\sqrt a + \sqrt a - 2a - 1\\
= 2\sqrt a \left( {1 - \sqrt a } \right) - \left( {1 - \sqrt a } \right)\\
= \left( {1 - \sqrt a } \right)\left( {2\sqrt a - 1} \right)\\
d)4a - 4\sqrt a - 1\\
= 4a - 4\sqrt a + 1 - 2\\
= {\left( {2\sqrt a + 1} \right)^2} - 2\\
= \left( {2\sqrt a + 1 + \sqrt 2 } \right)\left( {2\sqrt a + 1 - \sqrt 2 } \right)\\
e)a + 2\sqrt {ab} + \sqrt a + 2\sqrt b \\
= \sqrt a \left( {\sqrt a + 2\sqrt b } \right) + \left( {\sqrt a + 2\sqrt b } \right)\\
= \left( {\sqrt a + 2\sqrt b } \right)\left( {\sqrt a + 1} \right)\\
f)x - 3\sqrt x + 2\\
= \left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right)\\
g)x + 4\sqrt x + 3\\
= \left( {\sqrt x + 1} \right)\left( {\sqrt x + 3} \right)
\end{array}$
