1. Решите уравнение: 1) a) \(\frac{3x-x^2}{2} + \frac{2x^2-x}{6} = x\); 2) б) \(\frac{x^2}{x^2-x} = \frac{3x}{2-x}\); 3) б) \(\frac{y+4}{y+2} = \frac{2y-1}{y}\); 4) в) \(\frac{3y^2+y-24}{9-y^2} = -2\); 6) \(\frac{3x+1}{4} - \frac{7x-x^2}{10} = \frac{x^2-1}{8}\)

1) a) \(\frac{3x-x^2}{2} + \frac{2x^2-x}{6} = x\) Приведем к общему знаменателю 6: \(\frac{3(3x-x^2)}{6} + \frac{2x^2-x}{6} = \frac{6x}{6}\) \(9x - 3x^2 + 2x^2 - x = 6x\) \(-x^2 + 8x - 6x = 0\) \(-x^2 + 2x = 0\) \(x(-x + 2) = 0\) \(x_1 = 0, x_2 = 2\) Ответ: \(x_1 = 0, x_2 = 2\) 2) б) \(\frac{x^2}{x^2-x} = \frac{3x}{2-x}\) \(\frac{x^2}{x(x-1)} = \frac{3x}{2-x}\) \(\frac{x}{x-1} = \frac{3x}{2-x}\) \(x(2-x) = 3x(x-1)\) \(2x - x^2 = 3x^2 - 3x\) \(4x^2 - 5x = 0\) \(x(4x - 5) = 0\) \(x_1 = 0, x_2 = \frac{5}{4}\) Но \(x
eq 0\), так как на него делить нельзя, значит: Ответ: \(x = \frac{5}{4}\) 3) б) \(\frac{y+4}{y+2} = \frac{2y-1}{y}\) \(y(y+4) = (2y-1)(y+2)\) \(y^2 + 4y = 2y^2 + 4y - y - 2\) \(y^2 - y - 2 = 0\) \(D = (-1)^2 - 4*1*(-2) = 1 + 8 = 9\) \(y_1 = \frac{1 + 3}{2} = 2\) \(y_2 = \frac{1 - 3}{2} = -1\) Ответ: \(y_1 = 2, y_2 = -1\) 4) в) \(\frac{3y^2+y-24}{9-y^2} = -2\) \(3y^2 + y - 24 = -2(9 - y^2)\) \(3y^2 + y - 24 = -18 + 2y^2\) \(y^2 + y - 6 = 0\) \(D = 1 - 4*1*(-6) = 1 + 24 = 25\) \(y_1 = \frac{-1 + 5}{2} = 2\) \(y_2 = \frac{-1 - 5}{2} = -3\) Но \(y
eq 3\) и \(y
eq -3\), значит y = 2 Ответ: \(y = 2\) 6) \(\frac{3x+1}{4} - \frac{7x-x^2}{10} = \frac{x^2-1}{8}\) Приведём к общему знаменателю 40: \(\frac{10(3x+1)}{40} - \frac{4(7x-x^2)}{40} = \frac{5(x^2-1)}{40}\) \(30x + 10 - 28x + 4x^2 = 5x^2 - 5\) \(x^2 - 2x - 15 = 0\) \(D = 4 - 4*1*(-15) = 4 + 60 = 64\) \(x_1 = \frac{2 + 8}{2} = 5\) \(x_2 = \frac{2 - 8}{2} = -3\) Ответ: \(x_1 = 5, x_2 = -3\)