591. Решите уравнение: г) $$\frac{10}{2x-3}=x-1$$;

Решим уравнение:

$$\frac{10}{2x-3}=x-1$$

$$10 = (x-1)(2x-3)$$

$$10 = 2x^2 - 3x - 2x + 3$$

$$2x^2 - 5x - 7 = 0$$

По теореме Виета:

D = $$b^2 - 4ac = 25 - 4 \times 2 \times (-7) = 25 + 56 = 81$$

x1 = $$\frac{-b + \sqrt{D}}{2a} = \frac{5 + \sqrt{81}}{4} = \frac{5 + 9}{4} = \frac{14}{4} = 3.5$$

x2 = $$\frac{-b - \sqrt{D}}{2a} = \frac{5 - \sqrt{81}}{4} = \frac{5 - 9}{4} = \frac{-4}{4} = -1$$

Проверим:

$$\frac{10}{2 \times 3.5 -3} = \frac{10}{7-3} = \frac{10}{4} = 2.5$$

$$3.5 - 1 = 2.5$$

$$\frac{10}{2 \times (-1) -3} = \frac{10}{-2-3} = \frac{10}{-5} = -2$$

$$-1 -1 = -2$$

Ответ: 3.5; -1