4) б) \frac{2x^2-4x+3}{4x^2+3x-6} = 3;

4) б) Решим уравнение: $$ \frac{2x^2-4x+3}{4x^2+3x-6} = 3 $$

$$ 2x^2 - 4x + 3 = 3(4x^2 + 3x - 6) $$

$$ 2x^2 - 4x + 3 = 12x^2 + 9x - 18 $$

$$ 10x^2 + 13x - 21 = 0 $$

$$ D = 13^2 - 4(10)(-21) = 169 + 840 = 1009 $$

$$ x_1 = \frac{-13 + \sqrt{1009}}{20} $$

$$ x_2 = \frac{-13 - \sqrt{1009}}{20} $$

Ответ: $$x_1 = \frac{-13 + \sqrt{1009}}{20}, \quad x_2 = \frac{-13 - \sqrt{1009}}{20} $$