532. Решите уравнение: a) 3x² - 7x + 4 = 0; б) 5x² - 8x + 3 = 0; в) 3x² - 13x + 14 = 0; г) 2y² - 9y + 10 = 0.

а) Для уравнения \(3x^2 - 7x + 4 = 0\), \(a = 3\), \(b = -7\), \(c = 4\). \(D = (-7)^2 - 4 cdot 3 cdot 4 = 49 - 48 = 1\). \(x_1 = \frac{-(-7) + \sqrt{1}}{2 cdot 3} = \frac{7 + 1}{6} = \frac{8}{6} = \frac{4}{3}\) \(x_2 = \frac{-(-7) - \sqrt{1}}{2 cdot 3} = \frac{7 - 1}{6} = \frac{6}{6} = 1\) б) Для уравнения \(5x^2 - 8x + 3 = 0\), \(a = 5\), \(b = -8\), \(c = 3\). \(D = (-8)^2 - 4 cdot 5 cdot 3 = 64 - 60 = 4\). \(x_1 = \frac{-(-8) + \sqrt{4}}{2 cdot 5} = \frac{8 + 2}{10} = \frac{10}{10} = 1\) \(x_2 = \frac{-(-8) - \sqrt{4}}{2 cdot 5} = \frac{8 - 2}{10} = \frac{6}{10} = \frac{3}{5}\) в) Для уравнения \(3x^2 - 13x + 14 = 0\), \(a = 3\), \(b = -13\), \(c = 14\). \(D = (-13)^2 - 4 cdot 3 cdot 14 = 169 - 168 = 1\). \(x_1 = \frac{-(-13) + \sqrt{1}}{2 cdot 3} = \frac{13 + 1}{6} = \frac{14}{6} = \frac{7}{3}\) \(x_2 = \frac{-(-13) - \sqrt{1}}{2 cdot 3} = \frac{13 - 1}{6} = \frac{12}{6} = 2\) г) Для уравнения \(2y^2 - 9y + 10 = 0\), \(a = 2\), \(b = -9\), \(c = 10\). \(D = (-9)^2 - 4 cdot 2 cdot 10 = 81 - 80 = 1\). \(y_1 = \frac{-(-9) + \sqrt{1}}{2 cdot 2} = \frac{9 + 1}{4} = \frac{10}{4} = \frac{5}{2}\) \(y_2 = \frac{-(-9) - \sqrt{1}}{2 cdot 2} = \frac{9 - 1}{4} = \frac{8}{4} = 2\) **Ответ:** а) x₁ = 4/3, x₂ = 1 б) x₁ = 1, x₂ = 3/5 в) x₁ = 7/3, x₂ = 2 г) y₁ = 5/2, y₂ = 2