Giải thích các bước giải:
\(\begin{array}{l}
1/DK:x \ne 4\\
\frac{{14}}{{3\left( {x - 4} \right)}} - \frac{{2 + x}}{{x - 4}} = \frac{{ - 3}}{{2\left( {x - 4} \right)}} - \frac{5}{6}\\
\to \frac{{28 - 12 - 6x + 9 + 5\left( {x - 4} \right)}}{{12\left( {x - 4} \right)}} = 0\\
\to - x + 5 = 0\\
\to x = 5\left( {TM} \right)\\
b.DK:x \ne \left\{ { - 1;3} \right\}\\
\frac{{3{x^2} - 8x - 3 - \left( {2{x^2} - 3x - 5} \right) + 7}}{{\left( {x + 1} \right)\left( {x - 3} \right)}} = 0\\
\to {x^2} - 5x + 9 = 0\left( {voli} \right)
\end{array}\)
⇒ Pt vô nghiệm
\(\begin{array}{l}
3/DK:x \ne \left\{ {0; \pm 5} \right\}\\
\frac{{x + 5}}{{x\left( {x - 5} \right)}} - \frac{{x + 25}}{{2\left( {x - 5} \right)\left( {x + 5} \right)}} = \frac{{x - 5}}{{2x\left( {x + 5} \right)}}\\
\to \frac{{2\left( {{x^2} + 10x + 25} \right) - {x^2} - 25x - {x^2} + 10x - 25}}{{2x\left( {x - 5} \right)\left( {x + 5} \right)}} = 0\\
\to 5x - 25 = 0\\
\to x = 5\left( l \right)
\end{array}\)
⇒ Vô nghiệm
\(\begin{array}{l}
4/DK:x \ne - 2\\
\frac{2}{{x + 2}} - \frac{{2{x^2} + 16}}{{\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}} = \frac{5}{{{x^2} - 2x + 4}}\\
\to \frac{{2{x^2} - 4x + 8 - 2{x^2} - 16 - 5x - 10}}{{\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}} = 0\\
\to - 9x - 18 = 0\\
\to x = - 2\left( l \right)
\end{array}\)
⇒ Vô nghiệm
