Для решения уравнения \[\frac{2x-3}{x-12} = \frac{6x}{x-5}\] необходимо выполнить следующие шаги:
\[(2x-3)(x-5) = 6x(x-12)\]
\[2x^2 - 10x - 3x + 15 = 6x^2 - 72x\]
\[2x^2 - 13x + 15 = 6x^2 - 72x\]
\[0 = 4x^2 - 59x - 15\]
\[D = b^2 - 4ac = (-59)^2 - 4 \cdot 4 \cdot (-15) = 3481 + 240 = 3721\]
\[x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{59 + \sqrt{3721}}{8} = \frac{59 + 61}{8} = \frac{120}{8} = 15\]
\[x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{59 - \sqrt{3721}}{8} = \frac{59 - 61}{8} = \frac{-2}{8} = -0.25\]
При \( x = 15 \):
\[\frac{2(15)-3}{15-12} = \frac{30-3}{3} = \frac{27}{3} = 9\]
\[\frac{6(15)}{15-5} = \frac{90}{10} = 9\]
При \( x = -0.25 \):
\[\frac{2(-0.25)-3}{-0.25-12} = \frac{-0.5-3}{-12.25} = \frac{-3.5}{-12.25} = \frac{3.5}{12.25} = \frac{350}{1225} = \frac{2}{7}\]
\[\frac{6(-0.25)}{-0.25-5} = \frac{-1.5}{-5.25} = \frac{1.5}{5.25} = \frac{150}{525} = \frac{2}{7}\]
Оба корня подходят.
Ответ: x = 15; x = -0.25