Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 1334

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\[1)\ 2x^{- 2} + 4x^{- 1} + 3 = 0\ \ \ \ \ | \bullet x^{2}\]

\[2x^{0} + 4x^{1} + 3x^{2} = 0\]

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

\[D = 16 - 24 = - 8 < 0\]

\[корней\ нет.\]

\[Ответ:\ \ корней\ нет.\]

\[2)\ \left( x^{2} - x \right)^{2} + 12 = 8\left( x^{2} - x \right)\]

\[\left( x^{2} - x \right)^{2} - 8\left( x^{2} - x \right) + 12 = 0\]

\[y = \left( x^{2} - x \right):\]

\[y^{2} - 8y + 12 = 0\]

\[D = 64 - 48 = 16\]

\[y_{1} = \frac{8 - 4}{2} = 2;\text{\ \ }\]

\[y_{2} = \frac{8 + 4}{2} = 6.\]

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

\[x^{2} - x - 2 = 0\]

\[D = 1 + 8 = 9\]

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

\[x_{2} = \frac{1 + 3}{2} = 2.\]

\[2)\ x^{2} - x = 6\]

\[x^{2} - x - 6 = 0\]

\[D = 1 + 24 = 25\]

\[x_{3} = \frac{1 - 5}{2} = - 2;\]

\[x_{4} = \frac{1 + 5}{2} = 3.\]

\[Ответ:\ \ \]

\[x_{1} = - 1;\ \ x_{2,3} = \pm 2;\ \ x_{4} = 3.\]