Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\dfrac{5}{{3{x^2} - 12x}} = \dfrac{5}{{3x.\left( {x - 4} \right)}} = \dfrac{{5.2.\left( {x + 2} \right).\left( {x + 3} \right)}}{{3x.\left( {x - 4} \right).2.\left( {x + 2} \right).\left( {x + 3} \right)}}\\
= \dfrac{{10.\left( {x + 2} \right)\left( {x + 3} \right)}}{{6x.\left( {x - 4} \right)\left( {x + 2} \right)\left( {x + 3} \right)}} = \dfrac{{10.\left( {{x^2} + 5x + 6} \right)}}{{6x.\left( {x - 4} \right)\left( {x + 2} \right)\left( {x + 3} \right)}}\\
\dfrac{3}{{\left( {2x + 4} \right)\left( {x + 3} \right)}} = \dfrac{3}{{2.\left( {x + 2} \right)\left( {x + 3} \right)}} = \dfrac{{3.3x.\left( {x - 4} \right)}}{{2\left( {x + 2} \right)\left( {x + 3} \right).3x.\left( {x - 4} \right)}}\\
= \dfrac{{9x.\left( {x - 4} \right)}}{{6x.\left( {x - 4} \right)\left( {x + 2} \right)\left( {x + 3} \right)}} = \dfrac{{9{x^2} - 36x}}{{6x.\left( {x - 4} \right)\left( {x + 2} \right)\left( {x + 3} \right)}}
\end{array}\)
