ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 399

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Задание 399

\[\boxed{\mathbf{399}.}\]

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

\[x^{4} + 3x^{3} - x^{3} + x^{2} + x^{2} -\]

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

\[x^{4} + x^{2}(3x + 1) - x^{2}(x - 1) -\]

\[- x(3x + 1) + (3x + 1) = 0\]

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

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

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

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

\[D = 1 - 4 < 0\]

\[корней\ нет.\]

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

\[D = 9 - 4 = 5\]

\[x = \frac{- 3 \pm \sqrt{5}}{2}.\]

\[Ответ:x = \frac{- 3 \pm \sqrt{5}}{2}.\]

\[2)\ 2x^{4} + x^{3} - 10x^{2} - x + 2 = 0\]

\[Делители:\ \pm 1;\ \pm 2;\ \pm \frac{1}{2}.\]

\[2\] \[1\] \[- 10\] \[- 1\] \[2\]
\[2\] \[2\] \[5\] \[0\] \[- 1\] \[0\]
\[- 0,5\] \[2\] \[4\] \[- 2\] \[0\]

\[P(x) = (x - 2)(x + 0,5)\left( 2x^{2} + 4x - 2 \right) = 0\]

\[2x^{2} + 4x - 2 = 0\ \ \ |\ :2\]

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

\[D_{1} = 1 + 1 = 2\]

\[x = - 1 \pm \sqrt{2}.\]

\[Ответ:x = - 0,5;2;\ - 1 \pm \sqrt{2}.\]

\[3)\ (x - 1)x(x + 1)(x + 2) = 24\]

\[\left( x^{2} - 1 \right)\left( x^{2} + 2x \right) - 24 = 0\]

\[x^{4} + 2x^{3} - x^{2} - 2x - 24 = 0\]

\[Делители:\ \pm 1;\ \pm 2;\ \pm 3;\ \pm 4;\ \]

\[\pm 6;\ \pm 8;\ \pm 12;\ \pm 24.\]

\[1\] \[2\] \[- 1\] \[- 2\] \[- 24\]
\[- 3\] \[1\] \[- 1\] \[2\] \[- 8\] \[0\]
\[2\] \[1\] \[1\] \[4\] \[0\]

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

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

\[D = 1 - 16 < 0\]

\[нет\ корней.\]

\[Ответ:x = - 3;\ \ 2.\]

\[x^{2} + 5x = - 7\]

\[x^{2} + 5x + 7 = 0\]

\[D = 25 - 28 < 0\]

\[нет\ корней.\]

\[Ответ:x = \frac{- 5 \pm \sqrt{13}}{2}.\]

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