Đáp án:
$\begin{array}{l}
a)A = \dfrac{1}{2}\sqrt {16x} + \dfrac{5}{2}\sqrt {\dfrac{{4x}}{{25}}} - \dfrac{1}{x}\sqrt {\dfrac{{{x^3}}}{4}} \\
= \dfrac{1}{2}.4\sqrt x + \dfrac{5}{2}.\dfrac{2}{5}.\sqrt x - \dfrac{1}{x}.\dfrac{{x\sqrt x }}{2}\\
= 2\sqrt x + \sqrt x - \dfrac{1}{2}\sqrt x \\
= \dfrac{5}{2}\sqrt x \\
b)B = \dfrac{1}{x}\sqrt {50{x^3}} + \dfrac{1}{3}\sqrt {162x} - \sqrt {\dfrac{x}{{\sqrt {81} }}} \\
= \dfrac{1}{x}.5x\sqrt {2x} + \dfrac{1}{3}.9\sqrt {2x} - \dfrac{{\sqrt x }}{3}\\
= 5\sqrt {2x} + 3\sqrt {2x} - \dfrac{{\sqrt x }}{3}\\
= 8\sqrt {2x} - \dfrac{{\sqrt x }}{3}\\
= \dfrac{{\left( {24\sqrt 2 - 1} \right)\sqrt x }}{3}
\end{array}$
