Đáp án:
5) \(x = - 1\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:x \ne 5\\
\frac{{4x - 3}}{{x - 5}} = \frac{{29}}{3}\\
\to 12x - 9 = 29x - 145\\
\to 17x = 136\\
\to x = 8\left( {TM} \right)\\
2)DK:x \ne \frac{5}{3}\\
\frac{{4x - 5}}{{5 - 3x}} = 2\\
\to 4x - 5 = 10 - 6x\\
\to 10x = 15\\
\to x = \frac{3}{2}\left( {TM} \right)\\
3)DK:x \ne 1\\
\frac{{4x - 5}}{{x - 1}} = \frac{{2 + x}}{{x - 1}}\\
\to 4x - 5 = 2 + x\\
\to 3x = 7\\
\to x = \frac{7}{3}\left( {TM} \right)\\
4)DK:x \ne \left\{ { - 5;0} \right\}\\
\frac{{2x + 5}}{{2x}} - \frac{x}{{x + 5}} = 0\\
\to 2{x^2} + 15x + 25 - 2{x^2} = 0\\
\to 15x = - 25\\
\to x = - \frac{5}{3}\left( {TM} \right)\\
5)DK:x \ne \pm \frac{1}{3}\\
\frac{{12}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}} = \frac{{1 - 6x + 9{x^2} - 1 - 6x - 9{x^2}}}{{\left( {1 - 3x} \right)\left( {1 + 3x} \right)}}\\
\to - 12x = 12\\
\to x = - 1\left( {TM} \right)
\end{array}\)
