Đáp án:
\[\left[ \begin{array}{l}
x = 1\\
x = 5
\end{array} \right.\]
Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ne \pm 2\\
x \ne - \frac{3}{2}
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\frac{4}{{2{x^3} + 3{x^2} - 8x - 12}} - \frac{1}{{{x^2} - 4}} - \frac{4}{{2{x^2} + 7x + 6}} + \frac{1}{{2x + 3}} = 0\\
\Leftrightarrow \frac{4}{{{x^2}\left( {2x + 3} \right) - 4\left( {2x + 4} \right)}} - \frac{1}{{{x^2} - 4}} - \frac{4}{{x\left( {2x + 3} \right) + 2\left( {2x + 3} \right)}} + \frac{1}{{2x + 3}} = 0\\
\Leftrightarrow \frac{4}{{\left( {{x^2} - 4} \right)\left( {2x + 3} \right)}} - \frac{1}{{{x^2} - 4}} - \frac{4}{{\left( {x + 2} \right)\left( {2x + 3} \right)}} + \frac{1}{{2x + 3}} = 0\\
\Leftrightarrow \frac{{4 - \left( {2x + 3} \right).1 - 4.\left( {x - 2} \right) + \left( {{x^2} - 4} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {2x + 3} \right)}} = 0\\
\Leftrightarrow \frac{{4 - 2x - 3 - 4x + 8 + {x^2} - 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)\left( {2x + 3} \right)}} = 0\\
\Leftrightarrow {x^2} - 6x + 5 = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {x - 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 5
\end{array} \right.
\end{array}\)
