ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 751

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

\[Замена\ \ a = \left( \sqrt{2 + \sqrt{3}} \right)^{x};\ \ \]

\[b = \left( \sqrt{2 - \sqrt{3}} \right)^{x}:\]

\[a \cdot b =\]

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

\[= \left( \sqrt{4 - 3} \right)^{x} = \left( \sqrt{1} \right)^{x} = 1^{x} = 1.\]

\[\left\{ \begin{matrix} a + b = 4 \\ a \cdot b = 1\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} a = 4 - b\ \ \ \ \ \ \\ (4 - b)b = 1 \\ \end{matrix} \right.\ \]

\[4b - b^{2} - 1 = 0\]

\[b^{2} - 4b + 1 = 0\]

\[D_{1} = 4 - 1 = 3\]

\[b = 2 \pm \sqrt{3}.\]

\[a_{1} = 4 - 2 - \sqrt{3} = 2 - \sqrt{3};\]

\[a_{2} = 4 - 2 + \sqrt{3} = 2 + \sqrt{3}.\]

\[1)\ \ \left\{ \begin{matrix} \left( \sqrt{2 + \sqrt{3}} \right)^{x} = 2 + 2\sqrt{3} \\ \left( \sqrt{2 - \sqrt{3}} \right)^{x} = 2 - 2\sqrt{3} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

\[\frac{x}{2} = 1\]

\[x = 2.\]

\[2)\ \left\{ \begin{matrix} \left( \sqrt{2 + \sqrt{3}} \right)^{x} = 2 - 2\sqrt{3} \\ \left( \sqrt{2 - \sqrt{3}} \right)^{x} = 2 + 2\sqrt{3} \\ \end{matrix} \right.\ \]

\[\frac{x}{2} = - 1\]

\[x = - 2.\]

\[Ответ:x = \pm 2.\]