2. Найдите корни уравнения: 1) a) \(\frac{x-7}{x-2} = \frac{x+4}{x+2}\); 2) б) \(\frac{3y-3}{3y-2} + \frac{6+2y}{3y+2} = 2\); 3) a) \(\frac{4}{y-2} - \frac{3-y}{y^2-2y} = \frac{18}{y^2-4}\);

1) a) \(\frac{x-7}{x-2} = \frac{x+4}{x+2}\) \((x-7)(x+2) = (x+4)(x-2)\) \(x^2 + 2x - 7x - 14 = x^2 - 2x + 4x - 8\) \(x^2 - 5x - 14 = x^2 + 2x - 8\) \(-5x - 2x = 14 - 8\) \(-7x = 6\) \(x = -\frac{6}{7}\) Ответ: \(x = -\frac{6}{7}\) 2) б) \(\frac{3y-3}{3y-2} + \frac{6+2y}{3y+2} = 2\) \(\frac{(3y-3)(3y+2) + (6+2y)(3y-2)}{(3y-2)(3y+2)} = 2\) \(\frac{9y^2 + 6y - 9y - 6 + 18y - 12 + 6y^2 - 4y}{9y^2 - 4} = 2\) \(15y^2 + 11y - 18 = 2(9y^2 - 4)\) \(15y^2 + 11y - 18 = 18y^2 - 8\) \(3y^2 - 11y + 10 = 0\) \(D = (-11)^2 - 4*3*10 = 121 - 120 = 1\) \(y_1 = \frac{11 + 1}{6} = 2\) \(y_2 = \frac{11 - 1}{6} = \frac{5}{3}\) Но при y = 2 знаменатель первой дроби равен 4, поэтому y
eq 2 Ответ: \(y = \frac{5}{3}\) 3) a) \(\frac{4}{y-2} - \frac{3-y}{y^2-2y} = \frac{18}{y^2-4}\) \(\frac{4}{y-2} - \frac{3-y}{y(y-2)} = \frac{18}{(y-2)(y+2)}\) Умножим все части на \(y(y-2)(y+2)\): \(4y(y+2) - (3-y)(y+2) = 18y\) \(4y^2 + 8y - (3y + 6 - y^2 - 2y) = 18y\) \(4y^2 + 8y - 3y - 6 + y^2 + 2y = 18y\) \(5y^2 + 7y - 6 = 18y\) \(5y^2 - 11y - 6 = 0\) \(D = (-11)^2 - 4*5*(-6) = 121 + 120 = 241\) \(y_1 = \frac{11 + \sqrt{241}}{10}\) \(y_2 = \frac{11 - \sqrt{241}}{10}\) Ответ: \(y_1 = \frac{11 + \sqrt{241}}{10}, y_2 = \frac{11 - \sqrt{241}}{10}\)