Đáp án:
\[x = \frac{7}{5}\]
Giải thích các bước giải:
ĐKXĐ: \(x \ne \frac{1}{2}\)
Ta có:
\(\begin{array}{l}
M = \frac{{5x - 5}}{3} - \frac{{4\left( {x - 1} \right)\left( {x + 1} \right)}}{{2x - 1}}\\
\Leftrightarrow M = \frac{{\left( {5x - 5} \right)\left( {2x - 1} \right) - 12\left( {x - 1} \right)\left( {x + 1} \right)}}{{3\left( {2x - 1} \right)}}\\
\Leftrightarrow M = \frac{{10{x^2} - 15x + 5 - 12\left( {{x^2} - 1} \right)}}{{3\left( {2x - 1} \right)}}\\
\Leftrightarrow M = \frac{{ - 2{x^2} - 15x + 17}}{{6x - 3}}\\
\Leftrightarrow M = \frac{{2{x^2} + 15x - 17}}{{3 - 6x}}\\
N = \frac{{\left( {x - 3} \right)\left( {2x - 1} \right)}}{{6 - 12x}} - 1 = \frac{{2{x^2} - 7x + 3 - 6 + 12x}}{{6 - 12x}} = \frac{{2{x^2} + 5x - 3}}{{6 - 12x}}\\
M = 2N\\
\Leftrightarrow \frac{{2{x^2} + 15x - 17}}{{3 - 6x}} = 2.\frac{{2{x^2} + 5x - 3}}{{6 - 12x}}\\
\Leftrightarrow \frac{{2{x^2} + 15x - 17}}{{3 - 6x}} = \frac{{2{x^2} + 5x - 3}}{{3 - 6x}}\\
\Leftrightarrow 2{x^2} + 15x - 17 = 2{x^2} + 5x - 3\\
\Leftrightarrow 10x = 14\\
\Leftrightarrow x = \frac{7}{5}
\end{array}\)
