Đáp án:
$\begin{array}{l}
A = \frac{{4x + 5}}{{5y}} + \frac{x}{{5y}} - \frac{1}{y}\left( {y \ne 0} \right)\\
= \frac{{4x + 5}}{{5y}} + \frac{x}{{5y}} - \frac{5}{{5y}}\\
= \frac{{4x + 5 + x - 5}}{{5y}}\\
= \frac{{5x}}{{5y}} = \frac{x}{y}\\
B = \left( {\frac{2}{{x - 3}} - \frac{1}{{{x^2} - 9}}} \right):\frac{2}{{3{x^2} - 27}}\left( {x \ne \pm 3} \right)\\
= \left( {\frac{2}{{x - 3}} - \frac{1}{{\left( {x - 3} \right)\left( {x + 3} \right)}}} \right):\frac{2}{{3\left( {{x^2} - 9} \right)}}\\
= \frac{{2\left( {x + 3} \right) - 1}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\frac{{3\left( {x - 3} \right)\left( {x + 3} \right)}}{2}\\
= \frac{{3\left( {2x + 6 - 1} \right)}}{2}\\
= \frac{{6x + 15}}{2}
\end{array}$
