Đáp án:
a. x=2
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne \pm 1\\
\frac{{6 + 5\left( {{x^2} - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = \frac{{\left( {8x - 1} \right)\left( {x - 1} \right) + \left( {12x - 1} \right)\left( {x + 1} \right)}}{{4\left( {x - 1} \right)\left( {x + 1} \right)}}\\
\to 24 + 20{x^2} - 20 = 8{x^2} - 9x + 1 + 12{x^2} + 11x - 1\\
\to 2x = 4\\
\to x = 2\left( {TM} \right)
\end{array}\)
\(\begin{array}{l}
b.DK:x \ne \left\{ {1;2;3} \right\}\\
\frac{{x + 4}}{{\left( {x - 2} \right)\left( {x - 1} \right)}} - \frac{{x + 1}}{{\left( {x - 1} \right)\left( {x - 3} \right)}} - \frac{{2x + 5}}{{\left( {x - 1} \right)\left( {x - 3} \right)}} = 0\\
\to \frac{{x + 4}}{{\left( {x - 2} \right)\left( {x - 1} \right)}} - \frac{{x + 1 - 2x - 5}}{{\left( {x - 1} \right)\left( {x - 3} \right)}} = 0\\
\to \frac{{x + 4}}{{\left( {x - 2} \right)\left( {x - 1} \right)}} - \frac{{ - x - 4}}{{\left( {x - 1} \right)\left( {x - 3} \right)}} = 0\\
\to \frac{{x + 4}}{{\left( {x - 2} \right)\left( {x - 1} \right)}} + \frac{{x + 4}}{{\left( {x - 1} \right)\left( {x - 3} \right)}} = 0\\
\to \frac{{\left( {x + 4} \right)\left( {x - 3} \right) + \left( {x + 4} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x - 1} \right)\left( {x - 3} \right)}} = 0\\
\to {x^2} + x - 12 + {x^2} + 2x - 8 = 0\\
\to 2{x^2} + 3x - 20 = 0\\
\to 2{x^2} + 8x - 5x - 20 = 0\\
\to 2x\left( {x + 4} \right) - 5\left( {x + 4} \right) = 0\\
\to \left( {x + 4} \right)\left( {2x - 5} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 4\\
x = \frac{5}{3}
\end{array} \right.
\end{array}\)
