Đáp án:
$\begin{array}{l}
1){\left( {\sqrt x + 3} \right)^2} = x + 6\sqrt x + 9\\
2){\left( {1 + 2\sqrt x } \right)^2} = 1 + 4\sqrt x + 4x\\
3){\left( {\sqrt x - 2} \right)^2} = x - 4\sqrt x + 4\\
4){\left( {3\sqrt x - 1} \right)^2} = 9x - 6\sqrt x + 1\\
5){\left( {\sqrt x + 3\sqrt y } \right)^2} = x + 6\sqrt {xy} + 9y\\
6)\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) = x - 1\\
7)\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right) = x - 9\\
8){\left( {\sqrt x + 2} \right)^3}\\
= {\left( {\sqrt x } \right)^3} + 3.{\left( {\sqrt x } \right)^2}.2 + 3.\sqrt x .4 + 8\\
= x\sqrt x + 6x + 12\sqrt x + 8\\
9)\left( {2\sqrt x - \sqrt y } \right)\left( {2\sqrt x + \sqrt y } \right)\\
= 4x - y\\
10)x - 5 = \left( {\sqrt x - \sqrt 5 } \right)\left( {\sqrt x + \sqrt 5 } \right)\\
11)x - 16 = \left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)\\
12)x\sqrt x + 1\\
= {\left( {\sqrt x } \right)^3} + 1\\
= \left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)\\
13)y\sqrt y + 8\\
= {\left( {\sqrt y } \right)^3} + {2^3}\\
= \left( {\sqrt y + 2} \right)\left( {y - 2\sqrt y + 4} \right)\\
14){\left( {2\sqrt x - 1} \right)^3}\\
= 8x\sqrt x - 12x + 6\sqrt x - 1
\end{array}$
