Решите уравнения из группы C: 1) $(2x^2 + 7x - 8)(2x^2 + 7x - 3) = 314$, 2) $2y^3 + 2y^2 - (y + 1)^2 = 0$, 3) $6x^3 - 31x^2 - 31x + 6 = 0$, 4) $3x^2(x - 1)(x + 1) - 10x^2 + 4 = 0$

**Решение уравнений группы C:** 1) $(2x^2 + 7x - 8)(2x^2 + 7x - 3) = 314$ * Замена: $t = 2x^2 + 7x$, получаем $(t - 8)(t - 3) = 314$ * $t^2 - 11t + 24 = 314$ * $t^2 - 11t - 290 = 0$ * $D = (-11)^2 - 4 cdot 1 cdot (-290) = 121 + 1160 = 1281$ * $t_{1,2} = rac{11 pm sqrt{1281}}{2}$ * Возвращаемся к замене: $2x^2 + 7x = rac{11 pm sqrt{1281}}{2}$ * $4x^2 + 14x - (11 pm sqrt{1281}) = 0$ * $x = rac{-7 pm sqrt{49 + 4(11 pm sqrt{1281})}}{4} = rac{-7 pm sqrt{93 pm 4sqrt{1281}}}{4}$. Это сложное выражение, но решение возможно. 2) $2y^3 + 2y^2 - (y + 1)^2 = 0$ * $2y^3 + 2y^2 - (y^2 + 2y + 1) = 0$ * $2y^3 + y^2 - 2y - 1 = 0$ * Сгруппируем: $y^2(2y + 1) - (2y + 1) = 0$ * $(2y + 1)(y^2 - 1) = 0$ * $(2y + 1)(y - 1)(y + 1) = 0$ * Ответ: $y_1 = -rac{1}{2}, y_2 = 1, y_3 = -1$ 3) $6x^3 - 31x^2 - 31x + 6 = 0$ * Уравнение симметричное. Разделим на $x^3$: $6 - rac{31}{x} - rac{31}{x^2} + rac{6}{x^3}=0$ * Сгруппируем: $6(x^3 + rac{1}{x^3}) - 31(x + rac{1}{x}) = 0$ * Замена: $t = x + rac{1}{x}$, тогда $t^3 = x^3 + 3x + rac{3}{x} + rac{1}{x^3} = x^3 + rac{1}{x^3} + 3(x + rac{1}{x})$, т.е. $x^3 + rac{1}{x^3} = t^3 - 3t$ * $6(t^3 - 3t) - 31t = 0 Rightarrow 6t^3 - 18t - 31t = 0$ * $6t^3 - 49t = 0 Rightarrow t(6t^2 - 49) = 0$ * $t = 0$ или $6t^2 - 49 = 0$ * 1) $t = x + rac{1}{x} = 0 Rightarrow x^2 + 1 = 0$. Корней нет. * 2) $6t^2 = 49 Rightarrow t = pmsqrt{rac{49}{6}} = pmrac{7}{sqrt{6}}$ * $x + rac{1}{x} = rac{7}{sqrt{6}} Rightarrow sqrt{6}x^2 - 7x + sqrt{6} = 0$. $D = 49 - 4sqrt{6}sqrt{6} = 49 - 24 = 25$. $x_{1,2} = rac{7 pm 5}{2sqrt{6}} = rac{12}{2sqrt{6}} = sqrt{6}$ or $x=rac{2}{2sqrt{6}}=rac{1}{sqrt{6}}$ * $x + rac{1}{x} = -rac{7}{sqrt{6}} Rightarrow sqrt{6}x^2 + 7x + sqrt{6} = 0$. $D = 49 - 24 = 25$. $x_{3,4} = rac{-7 pm 5}{2sqrt{6}} = rac{-12}{2sqrt{6}} = -sqrt{6}$ or $x=rac{-2}{2sqrt{6}}=-rac{1}{sqrt{6}}$ * Ответ: $x_1 = sqrt{6}, x_2 = rac{1}{sqrt{6}}, x_3 = -sqrt{6}, x_4 = -rac{1}{sqrt{6}}$ 4) $3x^2(x - 1)(x + 1) - 10x^2 + 4 = 0$ * $3x^2(x^2 - 1) - 10x^2 + 4 = 0$ * $3x^4 - 3x^2 - 10x^2 + 4 = 0$ * $3x^4 - 13x^2 + 4 = 0$ * Замена: $t = x^2$, получаем $3t^2 - 13t + 4 = 0$ * $D = (-13)^2 - 4 cdot 3 cdot 4 = 169 - 48 = 121$ * $t_{1,2} = rac{13 pm sqrt{121}}{6} = rac{13 pm 11}{6}$ * $t_1 = rac{24}{6} = 4$ * $t_2 = rac{2}{6} = rac{1}{3}$ * Возвращаемся к замене: $x^2 = 4$ или $x^2 = rac{1}{3}$ * $x^2 = 4 Rightarrow x = pm 2$ * $x^2 = rac{1}{3} Rightarrow x = pm rac{1}{sqrt{3}} = pm rac{sqrt{3}}{3}$ * Ответ: $x_1 = 2, x_2 = -2, x_3 = rac{sqrt{3}}{3}, x_4 = -rac{sqrt{3}}{3}$