ГДЗ по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 18

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\[\textbf{а)}\sin x\log_{11}\left( 4 - x^{2} \right) = 0\]

\[1)\ \left\{ \begin{matrix} \sin x = 0\ \ \ \ \\ 4 - x^{2} > 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = \pi n\ \ \ \ \ \ \ \ \\ - 2 < x < 2 \\ \end{matrix} \right.\ \]

\[x = 0.\]

\[2)\ \left\{ \begin{matrix} \log_{11}\left( 4 - x^{2} \right) = 0 \\ x \in R\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[4 - x^{2} = 1\]

\[x^{2} = 3\]

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

\[Ответ:x = - \sqrt{3};x = 0;x = \sqrt{3}.\]

\[\textbf{б)}\cos x\log_{12}\left( 9 - x^{2} \right) = 0\]

\[1)\ \left\{ \begin{matrix} \cos x = 0\ \ \ \\ 9 - x^{2} \geq 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = \frac{\pi}{2} + \pi n \\ - 3 < x < 3 \\ \end{matrix} \right.\ \]

\[x = \pm \frac{\pi}{2}.\]

\[2)\ \left\{ \begin{matrix} \log_{12}\left( 9 - x^{2} \right) = 0 \\ x \in R\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

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

\[x^{2} = 8\]

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

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

\[\textbf{в)}\ tgx\log_{13}\left( x^{2} - x - 6 \right) = 0\]

\[1)\ \left\{ \begin{matrix} tg\ x = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - x - 6 > 0 \\ \end{matrix} \right.\ \]

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

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

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

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

\[x < - 2;\ \ x > 3.\]

\[\left\{ \begin{matrix} x = \pi k;\ \ \ k \neq 0 \\ x < - 2;\ \ x > 3 \\ \end{matrix} \right.\ \]

\[2)\ \left\{ \begin{matrix} \log_{13}\left( x^{2} - x - 6 \right) = 0\ \ \\ - \frac{\pi}{2} + \pi n < x < \frac{\pi}{2} + \pi n \\ \end{matrix} \right.\ \]

\[x^{2} - x - 6 = 1\]

\[x^{2} - x - 7 = 0\]

\[D = 1 + 28 = 29\]

\[x = \frac{1 \pm \sqrt{29}}{2}.\]

\[Ответ:x = \pi k\ (k \neq 0);\]

\[\ \ x = \frac{1 \pm \sqrt{29}}{2}.\]

\[\textbf{г)}\ ctgx\log_{14}\left( x^{2} + x - 12 \right) = 0\]

\[1)\ \left\{ \begin{matrix} ctg\ x = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + x - 12 > 0 \\ \end{matrix} \right.\ \]

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

\[x_{1} + x_{2} = - 1;\ \ x_{1} \cdot x_{2} = - 12\]

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

\[(x + 4)(x - 3) > 0\]

\[x < - 4;\ \ \ x > 3.\]

\[\left\{ \begin{matrix} x = \frac{\pi}{2} + \pi n;\ \ n \neq 0;n \neq 1\ \\ x < - 4;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x > 3 \\ \end{matrix} \right.\ \]

\[2)\ \left\{ \begin{matrix} \log_{14}\left( x^{2} + x - 12 \right) = 0 \\ \pi k < x < \pi + \pi k\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x^{2} + x - 12 = 1\]

\[x^{2} + x - 13 = 0\]

\[D = 1 + 52 = 53\]

\[x = \frac{- 1 \pm \sqrt{53}}{2}.\]

\[Ответ:x = \frac{\pi}{2} + \pi n\ (n \neq 0;n \neq 1);\]

\[x = \frac{- 1 \pm \sqrt{53}}{2}.\]