Đáp án:
$\begin{array}{l}
2)A = \sqrt {125} - 4\sqrt {45} + 3\sqrt {20} - \sqrt {80} \\
= 5\sqrt 5 - 4.3\sqrt 5 + 3.2\sqrt 5 - 4\sqrt 5 \\
= 5\sqrt 5 - 12\sqrt 5 + 6\sqrt 5 - 4\sqrt 5 \\
= - 5\sqrt 5 \\
B = \dfrac{5}{{4 - \sqrt {11} }} + \dfrac{1}{{3 + \sqrt 7 }} - \dfrac{6}{{\sqrt 7 - 2}} - \dfrac{{\sqrt 7 - 5}}{2}\\
= \dfrac{{5.\left( {4 + \sqrt {11} } \right)}}{{16 - 11}} + \dfrac{{3 - \sqrt 7 }}{{9 - 7}} - \dfrac{{6.\left( {\sqrt 7 + 2} \right)}}{{7 - 4}} - \dfrac{{\sqrt 7 - 5}}{2}\\
= 4 + \sqrt {11} + \dfrac{{3 - \sqrt 7 }}{2} - 2\left( {\sqrt 7 + 2} \right) - \dfrac{{\sqrt 7 - 5}}{2}\\
= \sqrt {11} - 2\sqrt 7 + \dfrac{{8 - 2\sqrt 7 }}{2}\\
= 4 + \sqrt {11} - 3\sqrt 7 \\
3)\\
2\sqrt {2x} - 5\sqrt {8x} + 7\sqrt {18x} = 28\left( {dkxd:x \ge 0} \right)\\
\Rightarrow 2\sqrt {2x} - 10\sqrt {2x} + 7.3\sqrt {2x} = 28\\
\Rightarrow 13\sqrt {2x} = 28\\
\Rightarrow \sqrt {2x} = \dfrac{{28}}{{13}}\\
\Rightarrow x = \dfrac{{28.14}}{{169}} = \dfrac{{392}}{{169}}\left( {tmdk} \right)
\end{array}$
