Đáp án:
$\begin{array}{l}
b)A = \dfrac{{6{x^3}}}{{2x - y}}\sqrt {\dfrac{{5\left( {{y^2} - 4xy + 4{x^2}} \right)}}{{6{x^2}}}} \\
= \dfrac{{6{x^3}}}{{2x - y}}.\dfrac{{\sqrt 5 \sqrt {{{\left( {2x - y} \right)}^2}} }}{{\sqrt 6 .x}}\\
= \dfrac{{\sqrt 6 {x^2}.\sqrt 5 .\left( {2x - y} \right)}}{{2x - y}}\left( {do:2x > y} \right)\\
= \sqrt {30} {x^2}\\
b)Dkxd:x \ge - 2\\
2\sqrt {4x + 8} - \sqrt {9x + 18} - \sqrt {\dfrac{{x + 2}}{4}} = 2\\
\Leftrightarrow 2.2\sqrt {x + 2} - 3\sqrt {x + 2} - \dfrac{1}{2}\sqrt {x + 2} = 2\\
\Leftrightarrow 4\sqrt {x + 2} - 3\sqrt {x + 2} - \dfrac{1}{2}\sqrt {x + 2} = 2\\
\Leftrightarrow \dfrac{1}{2}\sqrt {x + 2} = 2\\
\Leftrightarrow \sqrt {x + 2} = 4\\
\Leftrightarrow x + 2 = 16\\
\Leftrightarrow x = 14\left( {tmdk} \right)\\
Vậy\,x = 14
\end{array}$
