ариант 3 • 1. Решите уравнение: a) 7x29x + 2 = 0; в) 7х2-28 = 0; 6) 5x2 = 12x; г) х² + 20x + 91 = 0.

Вариант 3

1. Решите уравнение: a) $$7x^2 - 9x + 2 = 0$$ $$D = b^2 - 4ac = (-9)^2 - 4 \cdot 7 \cdot 2 = 81 - 56 = 25$$ $$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{9 + \sqrt{25}}{2 \cdot 7} = \frac{9 + 5}{14} = \frac{14}{14} = 1$$ $$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{9 - \sqrt{25}}{2 \cdot 7} = \frac{9 - 5}{14} = \frac{4}{14} = \frac{2}{7}$$ б) $$5x^2 = 12x$$ $$5x^2 - 12x = 0$$ $$x(5x - 12) = 0$$ $$x_1 = 0$$ $$5x - 12 = 0$$ $$5x = 12$$ $$x_2 = \frac{12}{5} = 2.4$$ в) $$7x^2 - 28 = 0$$ $$7x^2 = 28$$ $$x^2 = \frac{28}{7} = 4$$ $$x_1 = \sqrt{4} = 2$$ $$x_2 = -\sqrt{4} = -2$$ г) $$x^2 + 20x + 91 = 0$$ $$D = b^2 - 4ac = 20^2 - 4 \cdot 1 \cdot 91 = 400 - 364 = 36$$ $$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-20 + \sqrt{36}}{2 \cdot 1} = \frac{-20 + 6}{2} = \frac{-14}{2} = -7$$ $$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-20 - \sqrt{36}}{2 \cdot 1} = \frac{-20 - 6}{2} = \frac{-26}{2} = -13$$ Ответ: a) $$x_1 = 1, x_2 = \frac{2}{7}$$; б) $$x_1 = 0, x_2 = 2.4$$; в) $$x_1 = 2, x_2 = -2$$; г) $$x_1 = -7, x_2 = -13$$