ГДЗ по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 19

\[\boxed{\mathbf{19.}}\]

\[1)\ \left\{ \begin{matrix} \cos{2x} - 3\cos x - 1 = 0 \\ \log_{\frac{1}{3}}(x - 2) + 2 \geq 0\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

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

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

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

\[\cos x = t;\ - 1 \leq t \leq 1:\]

\[2t^{2} - 3t - 2 = 0\]

\[D = 9 + 16 = 25\]

\[t_{1} = \frac{3 + 5}{4} = 2\ (не\ подходит);\]

\[t_{2} = \frac{3 - 5}{4} = - \frac{1}{2}.\]

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

\[x = \frac{2\pi}{3} + 2\pi n.\]

\[\left\{ \begin{matrix} \log_{\frac{1}{3}}{(x - 2)} \geq - 2 \\ x - 2 > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x - 2 \leq \left( \frac{1}{3} \right)^{- 2}\]

\[x - 2 \leq 3^{2}\]

\[x - 2 \leq 9\]

\[x \leq 11.\]

\[\left\{ \begin{matrix} x \leq 11 \\ x > 2\ \ \ \\ \end{matrix} \right.\ \]

\[Данному\ неравенству\ \]

\[удовлетворяют\ корни:\]

\[\frac{2\pi}{3};\ \frac{4\pi}{3};\ \frac{8\pi}{3};\ \frac{10\pi}{3}.\]

\[2)\ \left\{ \begin{matrix} \log_{\frac{1}{3}}{(x - 2)} + 2 = 0 \\ x \in R\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\log_{\frac{1}{3}}(x - 2) = - 2\]

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

\[x - 2 = 9\]

\[x = 11.\]

\[Ответ:x = 11;x =\]

\[= \frac{2\pi}{3};\ \frac{4\pi}{3};\ \frac{8\pi}{3};\ \frac{10\pi}{3}.\]

\[1)\ \left\{ \begin{matrix} \cos{2x} + 7\cos x + 4 = 0 \\ \log_{\frac{1}{2}}{(x - 3)} + 1 \geq 0\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\cos{2x} + 7\cos x + 4 = 0\]

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

\[2cos^{2}x + 7\cos x + 3 = 0\]

\[\cos x = t;\ - 1 \leq t \leq 1:\]

\[2t^{2} + 7t + 3 = 0\]

\[D = 49 - 24 = 25\]

\[t_{1} = \frac{- 7 - 5}{4} = - 3\ (не\ подходит);\]

\[t_{2} = \frac{- 7 + 5}{4} = - \frac{1}{2}.\]

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

\[x = \frac{2\pi}{3} + 2\pi n.\]

\[\log_{\frac{1}{2}}(x - 3) + 1 \geq 0\]

\[x - 3 \leq 2\]

\[x \leq 5.\]

\[\left\{ \begin{matrix} x \leq 5 \\ x > 3 \\ \end{matrix} \right.\ \]

\[Ни\ один\ из\ корней\ \frac{2\pi}{3} + 2\pi n\ не\ \]

\[удовлетворяет\ неравенству.\]

\[2)\ \left\{ \begin{matrix} \log_{\frac{1}{2}}{(x - 3)} + 1 = 0 \\ x \in R\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\log_{\frac{1}{2}}(x - 3) + 1 = 0\]

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

\[x - 3 = 2\]

\[x = 5.\]

\[Ответ:x = 5.\]

\[\textbf{в)}\ \left( 4^{4 - x} - 2^{x - 1} \right)\log_{2}x = 0\]

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

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

\[2^{x} = t;\ \ t > 0:\]

\[\frac{4^{4}}{t^{2}} - \frac{t}{2} = 0\]

\[\frac{4^{4}}{t^{2}} = \frac{t}{2}\]

\[2^{9} = t^{3}\]

\[t^{3} = \left( 2^{3} \right)^{3}\]

\[t = 2^{3}.\]

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

\[x = 3.\]

\[2)\ \left\{ \begin{matrix} \log_{2}x = 0 \\ x \in R\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\log_{2}x = 0\]

\[x = 1.\]

\[Ответ:x = 1;x = 3.\]

\[\textbf{г)}\ \left( 4^{5 - x} - 2^{x - 1} \right)\log_{3}x = 0\]

\[1)\ \left\{ \begin{matrix} 4^{5 - x} - 2^{x - 1} = 0 \\ x > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[4^{5} \cdot 4^{- x} - 2^{x} \cdot 2^{- 1} = 0\]

\[2^{x} = t;\ \ t > 0:\]

\[\frac{4^{5}}{t^{2}} - \frac{t}{2} = 0\]

\[\frac{4^{5}}{t^{2}} = \frac{t}{2}\]

\[t^{3} = 2^{11}\]

\[t = 2^{\frac{11}{3}}\]

\[2^{x} = 2^{\frac{11}{3}}\]

\[x = \frac{11}{3} = 3\frac{2}{3}.\]

\[2)\ \left\{ \begin{matrix} \log_{3}x = 0 \\ x \in R\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\log_{3}x = 0\]

\[x = 1.\]

\[Ответ:x = 1;\ \ x = 3\frac{2}{3}.\]