Đáp án:
e. \(\dfrac{{12y - 8x}}{{\left( {y - 2x} \right)\left( {y + 2x} \right)}}\)
Giải thích các bước giải:
\(\begin{array}{l}
d.DK:x \ne \pm 2y;x \ne 0;y \ne 0\\
\dfrac{{2x}}{{x\left( {x + 2y} \right)}} + \dfrac{y}{{y\left( {x - 2y} \right)}} + \dfrac{4}{{\left( {x - 2y} \right)\left( {x + 2y} \right)}}\\
= \dfrac{2}{{x + 2y}} + \dfrac{1}{{x - 2y}} + \dfrac{4}{{\left( {x - 2y} \right)\left( {x + 2y} \right)}}\\
= \dfrac{{2x - 4y + x + 2y + 4}}{{\left( {x - 2y} \right)\left( {x + 2y} \right)}}\\
= \dfrac{{3x - 2y + 4}}{{\left( {x - 2y} \right)\left( {x + 2y} \right)}}\\
e.DK:y \ne 2x;x \ne 0\\
\dfrac{{2x + y + 2x - y}}{{x\left( {2x + y} \right)}} + \dfrac{{8y}}{{\left( {y - 2x} \right)\left( {y + 2x} \right)}}\\
= \dfrac{{4x}}{{x\left( {2x + y} \right)}} + \dfrac{{8y}}{{\left( {y - 2x} \right)\left( {y + 2x} \right)}}\\
= \dfrac{{4\left( {y - 2x} \right) + 8y}}{{\left( {y - 2x} \right)\left( {y + 2x} \right)}}\\
= \dfrac{{12y - 8x}}{{\left( {y - 2x} \right)\left( {y + 2x} \right)}}
\end{array}\)
