Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 114

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\[1)\ x^{4} - 7x^{2} + 12 = 0\]

\[x^{2} = y \geq 0:\]

\[y^{2} - 7y + 12 = 0\]

\[y_{1} + y_{2} = 7;\ \ \ y_{1} \cdot y_{2} = 12\]

\[y_{1} = 3;\ \ y_{2} = 4.\]

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

\[x = \pm \sqrt{3}.\]

\[2)\ x^{2} = 4\]

\[x = \pm 2.\]

\[Ответ:x = \pm \sqrt{3};\ \ x = \pm 2.\]

\[2)\ x^{4} - 10x^{2} + 9 = 0\]

\[x^{2} = y \geq 0:\]

\[y^{2} - 10y + 9 = 0\]

\[D_{2} = 25 - 9 = 16\]

\[y_{1} = 5 + 4 = 9;\ \ y_{2} = 5 - 4 = 1.\]

\[1)\ x^{2} = 9\]

\[x = \pm 3.\]

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

\[x = \pm 1.\]

\[Ответ:x = \pm 1;\ \ x = \pm 3.\]

\[3)\ x^{4} + 2x^{2} - 15 = 0\]

\[x^{2} = y \geq 0:\]

\[y^{2} + 2y - 15 = 0\]

\[D_{1} = 1 + 15 = 16\]

\[y_{1} = - 1 + 4 = 3;\ \]

\[\ y_{2} = - 1 - 4 =\]

\[= - 5 < 0\ (не\ подходит).\]

\[x^{2} = 3\]

\[x = \pm \sqrt{3}.\]

\[Ответ:x = \pm \sqrt{3}.\]

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

\[x^{2} = y \geq 0:\]

\[y^{2} + y - 6 = 0\]

\[y_{1} + y_{2} = - 1;\ \ y_{1} \cdot y_{2} = - 6\]

\[y_{1} = - 3 < 0\ (не\ подходит);\]

\[y_{2} = 2.\]

\[x^{2} = 2\]

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

\[Ответ:x = \pm \sqrt{2}.\]