Giải thích các bước giải:
\(\begin{array}{l}
ĐK:x \ne \left\{ {\frac{1}{5};\frac{3}{5};\frac{{1 \pm \sqrt {241} }}{{10}}} \right\}\\
\frac{3}{{5x - 1}} + \frac{2}{{3 - 5x}} = \frac{4}{{12 + x - 5{x^2}}}\\
\to \left( {3 - 5x} \right)\left( {12 + x - 5{x^2}} \right) + 2\left( {12 + x - 5{x^2}} \right)\left( {5x - 1} \right) = 4\left( {5x - 1} \right)\left( {3 - 5x} \right)\\
\to 3\left( {36 + 3x - 15{x^2} - 60x - 5{x^2} + 25{x^3}} \right) + 2\left( {60x + 5{x^2} - 25{x^3} - 12 - x + 5{x^2}} \right)\\
= 4\left( { - 25{x^2} + 20x - 3} \right)\\
\to 25{x^3} + 60{x^2} - 133x + 96 = 0\\
\to x = - 3,979383665\left( {TM} \right)
\end{array}\)
