Đáp án:
b) x=-40
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ { - 1;1;3} \right\}\\
\dfrac{x}{{2\left( {x - 3} \right)}} - \dfrac{x}{{2\left( {x - 1} \right)}} = \dfrac{{2x}}{{\left( {x + 1} \right)\left( {x - 3} \right)}}\\
\to \dfrac{{x\left( {x - 1} \right)\left( {x + 1} \right) - x\left( {x - 3} \right)\left( {x + 1} \right) - 4x\left( {x - 1} \right)}}{{2\left( {x - 3} \right)\left( {x + 1} \right)\left( {x - 1} \right)}} = 0\\
\to x\left( {{x^2} - 1} \right) - x\left( {{x^2} - 2x - 3} \right) - 4{x^2} + 4x = 0\\
\to {x^3} - x - {x^3} + 2{x^2} + 3x - 4{x^2} + 4x = 0\\
\to - 2{x^2} + 6x = 0\\
\to - 2x\left( {x - 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\left( {TM} \right)\\
x = 3\left( l \right)
\end{array} \right.\\
b)DK:x \ne \pm 4\\
5 + \dfrac{{96}}{{{x^2} - 16}} = \dfrac{{2x - 1}}{{x + 4}} + \dfrac{{3x - 1}}{{x - 4}}\\
\to \dfrac{{5{x^2} - 80 - \left( {2x - 1} \right)\left( {x - 4} \right) - \left( {3x - 1} \right)\left( {x + 4} \right)}}{{\left( {x - 4} \right)\left( {x + 4} \right)}} = 0\\
\to 5{x^2} - 80 - 2{x^2} + 9x - 4 - 3{x^2} - 11x + 4 = 0\\
\to - 2x = 80\\
\to x = - 40
\end{array}\)
