Em xem lại đề câu cuối nhé, là 15 (không phải 45) em nhé
Giải thích các bước giải:
\(\begin{array}{l}
f)\,\dfrac{x}{{x\left( {x + 2y} \right)}} + \dfrac{1}{{x - 2y}} - \dfrac{{4y}}{{\left( {x - 2y} \right)\left( {x + 2y} \right)}}\\
= \dfrac{{x - 2y + x + 2y - 4y}}{{\left( {x - 2y} \right)\left( {x + 2y} \right)}} = \dfrac{{2\left( {x - 2y} \right)}}{{\left( {x - 2y} \right)\left( {x + 2y} \right)}} = \dfrac{2}{{x + 2y}}\\
g)\,\dfrac{x}{{y\left( {2y - x} \right)}} + \dfrac{{4y}}{{x\left( {x - y} \right)}}\\
= \dfrac{{{x^2}\left( {x - y} \right) + 4{y^2}\left( {2y - x} \right)}}{{xy\left( {2y - x} \right)\left( {x - y} \right)}}\\
= \dfrac{{{x^3} + 8{y^3} - {x^2}y - 4x{y^2}}}{{xy\left( {2y - x} \right)\left( {x - y} \right)}}\\
h)\,\dfrac{{\left( {x + 1} \right)\left( {x - 2} \right) + x\left( {{x^2} + 2x + 4} \right) - {x^3} - 16}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
= \dfrac{{{x^2} - x - 2 + {x^3} + 2{x^2} + 4x - {x^3} - 16}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
= \dfrac{{3{x^2} + 3x - 18}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}} = \dfrac{{3\left( {x - 2} \right)\left( {x + 3} \right)}}{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}\\
= \dfrac{{3\left( {x + 3} \right)}}{{{x^2} + 2x + 4}}\\
l)\,\dfrac{2}{{\left( {x - 3} \right)\left( {x - 5} \right)}} + \dfrac{3}{{\left( {x - 2} \right)\left( {x - 5} \right)}}\\
= \dfrac{{2\left( {x - 2} \right) + 3\left( {x - 3} \right)}}{{\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 5} \right)}}\\
= \dfrac{{5x - 13}}{{\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 5} \right)}}
\end{array}\)
