Đáp án:
$\begin{array}{l}
a)A = 4\sqrt {\dfrac{{25x}}{4}} - \dfrac{8}{3}\sqrt {\dfrac{{9x}}{4}} - \dfrac{4}{{3x}}\sqrt {\dfrac{{9{x^3}}}{{64}}} \\
= 4.\dfrac{5}{2}\sqrt x - \dfrac{8}{3}.\dfrac{3}{2}\sqrt x - \dfrac{4}{{3x}}.\dfrac{{3x\sqrt x }}{8}\\
= 10\sqrt x - 4\sqrt x - \dfrac{1}{2}\sqrt x \\
= \dfrac{{11}}{2}\sqrt x \\
b)B = \dfrac{y}{2} + \dfrac{3}{4}\sqrt {1 - 4y + 4{y^2}} - \dfrac{3}{2}\\
= \dfrac{y}{2} + \dfrac{3}{4}.\sqrt {{{\left( {1 - 2y} \right)}^2}} - \dfrac{3}{2}\\
= \dfrac{y}{2} + \dfrac{3}{4}.\left( {1 - 2y} \right) - \dfrac{3}{2}\left( {do:y \le \dfrac{1}{2}} \right)\\
= \dfrac{y}{2} + \dfrac{3}{4} - \dfrac{3}{2}y - \dfrac{3}{2}\\
= - y - \dfrac{3}{4}
\end{array}$
