544. Решите уравнение: B) 4x2-1/3 = x(10x-9); Г) 3/4 x² - 2/5 x = 4/5 x² + 3/4 .

B) $$\frac{4x^2 - 1}{3} = x(10x - 9)$$

$$4x^2 - 1 = 3(10x^2 - 9x)$$

$$4x^2 - 1 = 30x^2 - 27x$$

$$26x^2 - 27x + 1 = 0$$

$$D = (-27)^2 - 4 \cdot 26 \cdot 1 = 729 - 104 = 625$$

$$x_1 = \frac{27 + \sqrt{625}}{2 \cdot 26} = \frac{27 + 25}{52} = \frac{52}{52} = 1$$

$$x_2 = \frac{27 - \sqrt{625}}{2 \cdot 26} = \frac{27 - 25}{52} = \frac{2}{52} = \frac{1}{26}$$

Ответ: $$x_1 = 1$$, $$x_2 = \frac{1}{26}$$

Г) $$\frac{3}{4}x^2 - \frac{2}{5}x = \frac{4}{5}x^2 + \frac{3}{4}$$

$$\frac{3}{4}x^2 - \frac{4}{5}x^2 - \frac{2}{5}x - \frac{3}{4} = 0$$

$$\frac{15 - 16}{20}x^2 - \frac{2}{5}x - \frac{3}{4} = 0$$

$$- \frac{1}{20}x^2 - \frac{2}{5}x - \frac{3}{4} = 0$$

$$x^2 + 8x + 15 = 0$$

$$D = 8^2 - 4 \cdot 1 \cdot 15 = 64 - 60 = 4$$

$$x_1 = \frac{-8 + \sqrt{4}}{2 \cdot 1} = \frac{-8 + 2}{2} = \frac{-6}{2} = -3$$

$$x_2 = \frac{-8 - \sqrt{4}}{2 \cdot 1} = \frac{-8 - 2}{2} = \frac{-10}{2} = -5$$

Ответ: $$x_1 = -3$$, $$x_2 = -5$$