Đáp án:
\[x = \frac{{81 \pm \sqrt {929} }}{{32}}\]
Giải thích các bước giải:
ĐKXĐ: \(\left\{ \begin{array}{l}
x \ne 2\\
x \ne 4
\end{array} \right.\)
Ta có:
\(\begin{array}{l}
\frac{{x - 3}}{{x - 2}} - \frac{{x - 2}}{{x - 4}} = 3\frac{1}{5}\\
\Leftrightarrow \frac{{\left( {x - 3} \right)\left( {x - 4} \right) - {{\left( {x - 2} \right)}^2}}}{{\left( {x - 2} \right)\left( {x - 4} \right)}} = \frac{{16}}{5}\\
\Leftrightarrow \frac{{{x^2} - 7x + 12 - {x^2} + 4x - 4}}{{{x^2} - 6x + 8}} = \frac{{16}}{5}\\
\Leftrightarrow \frac{{ - 3x + 8}}{{{x^2} - 6x + 8}} = \frac{{16}}{5}\\
\Leftrightarrow 16{x^2} - 96x + 128 = - 15x + 40\\
\Leftrightarrow 16{x^2} - 81x + 88 = 0\\
\Leftrightarrow x = \frac{{81 \pm \sqrt {929} }}{{32}}
\end{array}\)
