Решите уравнение: 3x^3 - 2x^2 - x = 0

[latex]3x^3 - 2x^2 - x = 0 \\ x(3x^2-2x-1) = 0 \\ x_1 = 0 \\ 3x^2-2x-1 = 0 \\ D = (-2)^2 - 4 * (-1) * 3 = 4 -(-12) = 4+12 = 16 \\ x_2_,_3 = \frac{2б \sqrt{D} }{2*3} = \frac{2б4}{6} \\ x_2 = 1 \\ x_3 = - \frac{2}{6} = -\frac{1}{3} [/latex] Ответ: [latex] -\frac{1}{3}; 0; 1 [/latex]