Đáp án:
$\begin{array}{l}
1)a)\\
A = \sqrt {{{\left( {5 - 2\sqrt 2 } \right)}^2}} + \sqrt {{{\left( {7 - 2\sqrt 2 } \right)}^2}} + \dfrac{7}{{\sqrt 7 }}\\
= \left| {5 - 2\sqrt 2 } \right| + \left| {7 - 2\sqrt 2 } \right| + \sqrt 7 \\
= 2\sqrt 2 - 5 + 2\sqrt 2 - 7 + \sqrt 7 \\
= 4\sqrt 2 - 12 + \sqrt 7 \\
b)B = \dfrac{{15}}{{\sqrt 6 + 1}} + \dfrac{4}{{\sqrt 6 - 2}} - \dfrac{{12}}{{3 - \sqrt 6 }} - \sqrt 6 \\
= \dfrac{{15\left( {\sqrt 6 - 1} \right)}}{{6 - 1}} + \dfrac{{4\left( {\sqrt 6 + 2} \right)}}{{6 - 4}} - \dfrac{{12\left( {3 + \sqrt 6 } \right)}}{{9 - 6}} - \sqrt 6 \\
= 3\left( {\sqrt 6 - 1} \right) + 2\left( {\sqrt 6 + 2} \right) - 4\left( {3 + \sqrt 6 } \right) - \sqrt 6 \\
= 3\sqrt 6 - 3 + 2\sqrt 6 + 4 - 12 - 4\sqrt 6 - \sqrt 6 \\
= - 11\\
2)a)\\
Dkxd:x \ge - \dfrac{1}{2}\\
3\sqrt {8x + 4} - \dfrac{1}{3}\sqrt {18x + 9} \\
- \dfrac{1}{2}\sqrt {50x + 25} + \sqrt {\dfrac{{2x + 1}}{4}} = 6\\
\Rightarrow 3.2\sqrt {2x + 1} - \dfrac{1}{3}.3\sqrt {2x + 1} \\
- \dfrac{1}{2}.5\sqrt {2x + 1} + \dfrac{1}{2}\sqrt {2x + 1} = 6\\
\Rightarrow 6\sqrt {2x + 1} - \sqrt {2x + 1} - \dfrac{5}{2}\sqrt {2x + 1} + \dfrac{1}{2}\sqrt {2x + 1} = 6\\
\Rightarrow 3\sqrt {2x + 1} = 6\\
\Rightarrow \sqrt {2x + 1} = 2\\
\Rightarrow 2x + 1 = 4\\
\Rightarrow 2x = 3\\
\Rightarrow x = \dfrac{3}{2}\left( {tmdk} \right)\\
\text{Vậy}\,x = \dfrac{3}{2}
\end{array}$
