592. Найдите корни уравнения: з) $$\frac{x^2-5}{x-1}=\frac{7x+10}{9}$$.

Решим уравнение:

$$\frac{x^2-5}{x-1}=\frac{7x+10}{9}$$

$$9(x^2 - 5) = (7x+10)(x-1)$$

$$9x^2 - 45 = 7x^2 -7x + 10x - 10$$

$$2x^2 - 3x - 35 = 0$$

D = $$b^2 - 4ac = 9 - 4 \times 2 \times (-35) = 9 + 280 = 289$$

x1 = $$\frac{-b + \sqrt{D}}{2a} = \frac{3 + \sqrt{289}}{4} = \frac{3 + 17}{4} = \frac{20}{4} = 5$$

x2 = $$\frac{-b - \sqrt{D}}{2a} = \frac{3 - \sqrt{289}}{4} = \frac{3 - 17}{4} = \frac{-14}{4} = -3.5$$

Проверим:

$$\frac{5^2-5}{5-1}=\frac{7 \times 5+10}{9}$$

$$\frac{20}{4}=\frac{45}{9}$$

$$5 = 5$$

$$\frac{(-3.5)^2-5}{-3.5-1}=\frac{7 \times (-3.5)+10}{9}$$

$$\frac{12.25-5}{-4.5}=\frac{-24.5+10}{9}$$

$$\frac{7.25}{-4.5}=\frac{-14.5}{9}$$

$$-1.61 = -1.61$$

Ответ: 5; -3.5