giải phương trình 4/2x^3+3x^2-8x-12 – 1/x^2-4 -4/2x^2+7x+6 +1/2x+3=0 giúp vs ạ đag gấp lắm

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:

\(\begin{array}{l}
\frac{4}{{2{x^3} + 3{x^2} – 8x – 12}} – \frac{1}{{{x^2} – 4}} – \frac{4}{{2{x^2} + 7x + 6}} + \frac{1}{{2x + 3}} = 0\\
 \Leftrightarrow \frac{4}{{{x^2}\left( {2x + 3} \right) – 4\left( {2x + 4} \right)}} – \frac{1}{{{x^2} – 4}} – \frac{4}{{x\left( {2x + 3} \right) + 2\left( {2x + 3} \right)}} + \frac{1}{{2x + 3}} = 0\\
 \Leftrightarrow \frac{4}{{\left( {{x^2} – 4} \right)\left( {2x + 3} \right)}} – \frac{1}{{{x^2} – 4}} – \frac{4}{{\left( {x + 2} \right)\left( {2x + 3} \right)}} + \frac{1}{{2x + 3}} = 0\\
 \Leftrightarrow \frac{{4 – \left( {2x + 3} \right).1 – 4.\left( {x – 2} \right) + \left( {{x^2} – 4} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right)\left( {2x + 3} \right)}} = 0\\
 \Leftrightarrow \frac{{4 – 2x – 3 – 4x + 8 + {x^2} – 4}}{{\left( {x – 2} \right)\left( {x + 2} \right)\left( {2x + 3} \right)}} = 0\\
 \Leftrightarrow {x^2} – 6x + 5 = 0\\
 \Leftrightarrow \left( {x – 1} \right)\left( {x – 5} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 5
\end{array} \right.
\end{array}\)