ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 814

\[1)\ x^{3} - 3x^{2} + x = 3\]

\[x^{2}(x - 3) + (x - 3) = 0\]

\[\left( x^{2} + 1 \right)(x - 3) = 0\]

\[x - 3 = 0\]

\[x = 3.\]

\[Ответ:\ \ 3.\]

\[2)\ x^{3} - 3x^{2} - 4x + 12 = 0\]

\[x^{2}(x - 3) - 4(x - 3) = 0\]

\[\left( x^{2} - 4 \right)(x - 3) = 0\]

\[(x + 2)(x - 2)(x - 3) = 0\]

\[x_{1} = - 2;\text{\ \ \ }x_{2} = 2;\text{\ \ \ }x_{3} = 3.\]

\[Ответ:\ - 2;\ 2;\ 3.\]

\[3)\ x^{3} - 5x^{2} + 8x - 6 = 0\]

\[(x - 3)\left( x^{2} - 2x + 2 \right) = 0\]

\[D = 4 - 8 = - 4 < 0\]

\[x \in \varnothing.\]

\[Ответ:\ \ 3.\]

\[4)\ x^{4} - 3x^{3} - 2x^{2} - 6x - 8 = 0\]

\[(x + 1)\left( x^{3} - 4x^{2} + 2x - 8 \right) = 0\]

\[(x + 1)\left( x^{2}(x - 4) + 2(x - 4) \right) = 0\]

\[(x + 1)(x - 4)\left( x^{2} + 2 \right) = 0\]

\[(x + 1)(x - 4) = 0\]

\[x_{1} = - 1;\text{\ \ \ }x_{2} = 4.\]

\[Ответ:\ - 1;\ 4.\]

\[(x - 3)(x + 1)^{4} = 0\]

\[x_{1} = - 1;\text{\ \ \ }x_{2} = 3.\]

\[Ответ:\ - 1;\ 3.\]

\[6)\ 2x^{4} - 2x^{3} - 11x^{2} - x - 6 = 0\]

\[(x + 2)\left( 2x^{3} - 6x^{2} + x - 3 \right) = 0\]

\[(x + 2)\left( 2x^{2}(x - 3) + (x - 3) \right) = 0\]

\[(x + 2)(x - 3)\left( 2x^{2} + 1 \right) = 0\]

\[(x + 2)(x - 3) = 0\]

\[x_{1} = - 2;\text{\ \ \ }x_{2} = 3.\]

\[Ответ:\ - 2;\ 3.\]