Đáp án:
$\begin{array}{l}
\frac{{2x + y}}{{2{x^2} - xy}} + \frac{{16x}}{{{y^2} - 4{x^2}}} + \frac{{2x - y}}{{2{x^2} + xy}}\\
= \frac{{2x + y}}{{x\left( {2x - y} \right)}} + \frac{{16x}}{{\left( {y + 2x} \right)\left( {y - 2x} \right)}} + \frac{{2x - y}}{{x\left( {2x + y} \right)}}\\
= \frac{{\left( {2x + y} \right)\left( {2x + y} \right) - 16x.x + \left( {2x - y} \right)\left( {2x - y} \right)}}{{x.\left( {2x + y} \right)\left( {2x - y} \right)}}\\
= \frac{{{{\left( {2x + y} \right)}^2} - 16{x^2} + {{\left( {2x - y} \right)}^2}}}{{x.\left( {2x + y} \right)\left( {2x - y} \right)}}\\
= \frac{{4{x^2} + 4xy + {y^2} - 16{x^2} + 4{x^2} - 4xy + {y^2}}}{{x.\left( {2x + y} \right)\left( {2x - y} \right)}}\\
= \frac{{2{y^2} - 8{x^2}}}{{x.\left( {2x + y} \right)\left( {2x - y} \right)}}\\
= \frac{{ - 2\left( {4{x^2} - {y^2}} \right)}}{{x\left( {4{x^2} - {y^2}} \right)}}\\
= - \frac{2}{x}
\end{array}$
