ГДЗ по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 40

\[\boxed{\mathbf{40.}}\]

\[\textbf{а)}\ \sqrt{x^{2} - 5} + \sqrt[4]{x^{2} - 3} =\]

\[= \sqrt{x + 1} + \sqrt[4]{x + 3}\]

\[\left\{ \begin{matrix} \sqrt{x^{2} - 5} = \sqrt{x + 1} \\ \sqrt[4]{x^{2} - 3} = \sqrt[4]{x + 3} \\ x^{2} - 5 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 3 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x + 1 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x + 3 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x^{2} - 5 = x + 1 \\ x^{2} - 3 = x + 3 \\ x^{2} \geq 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} \geq 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \geq - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \geq - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x^{2} - x - 6 = 0 \\ x^{2} - x - 6 = 0 \\ x^{2} \geq 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} \geq 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \geq - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \geq - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x^{2} - x - 6 = 0 \\ x^{2} \geq 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \geq - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ }\]

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

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

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

\[x_{2} = - 2\ (не\ подходит).\]

\[Ответ:x = 3.\]

\[\textbf{б)}\ \sqrt[4]{x^{2} - x - 3} + \sqrt{x^{2} - x + 5} =\]

\[= \sqrt[4]{2x + 1} + \sqrt{2x + 9}\]

\[\left\{ \begin{matrix} \sqrt[4]{x^{2} - x - 3} = \sqrt[4]{2x + 1} \\ \sqrt{x^{2} - x + 5} = \sqrt{2x + 9} \\ x^{2} - x - 3 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - x + 5 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x + 1 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x + 9 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x^{2} - x - 3 = 2x + 1 \\ x^{2} - x + 5 = 2x + 9 \\ x^{2} - x - 3 \geq 0\ \ \ \ \ \ \ \ \ \ \\ x \geq - \frac{1}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x \geq - 4,5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x^{2} - 3x - 4 = 0\ \ \ \ \ \ \ \\ x^{2} - 3x - 4 = 0\ \ \ \ \ \ \ \ \\ x^{2} - x - 3 \geq 0\ \ \ \ \ \ \ \ \ \ \\ x \geq - \frac{1}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x \geq - 4,5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

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

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

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

\[x = 4 - решение\ системы.\]

\[Ответ:x = 4.\]

\[\textbf{в)}\ \sqrt{\sin x - 0,1} + \sqrt[4]{\sin x + 0,9} =\]

\[= \sqrt{\cos x + 0,9} + \sqrt[4]{\cos x + 1,9}\]

\[\left\{ \begin{matrix} \sqrt{\sin x - 0,1} = \sqrt{\cos x + 0,9}\ \\ \sqrt[4]{\sin x + 0,9} = \sqrt[4]{\cos x + 1,9} \\ \sin x - 0,1 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \cos x + 0,9 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \sin x + 0,9 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \sin x - 0,1 = \cos x + 0,9 \\ \sin x + 0,9 = \cos x + 1,9 \\ \sin x \geq - 0,1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \cos x \geq - 0,9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \sin x \geq - 0,9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \sin x - \cos x = 1 \\ \sin x - \cos x = 1 \\ \sin x \geq 0,1\ \ \ \ \ \ \ \ \ \ \ \\ \cos x \geq - 0,9\ \ \ \ \ \ \ \ \\ \sin x \geq - 0,9\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \sin x - \cos x = 1 \\ \sin x \geq 0,1\ \ \ \ \ \ \ \ \ \ \ \ \\ \cos x \geq - 0,9\ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\sin x - \cos x = 1\ \ \ \ |\ :\sqrt{2}\]

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

\[\cos\frac{\pi}{4}\sin x - \sin\frac{\pi}{4}\cos x = \frac{1}{\sqrt{2}}\]

\[\sin\left( x - \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}\]

\[x - \frac{\pi}{4} = \frac{\pi}{4} + 2\pi k\]

\[x = \frac{\pi}{2} + 2\pi k;\]

\[x - \frac{\pi}{4} = \frac{3\pi}{4} + 2\pi k\]

\[x = \pi + 2\pi k.\]

\[Ответ:x = \frac{\pi}{2} + 2\pi k.\]

\[\textbf{г)}\ \sqrt[4]{\text{tgx}} + \sqrt{tgx + 1} =\]

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

\[\left\{ \begin{matrix} \sqrt[4]{\text{tgx}} = \sqrt[4]{2 - ctgx}\text{\ \ \ \ \ \ \ } \\ \sqrt{tgx + 1} = \sqrt{3 - ctgx} \\ tgx \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2 - ctgx \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ tgx + 1 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3 - ctgx \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} tgx = 2 - ctgx\ \ \ \ \ \ \ \ \\ tgx + 1 = 3 - ctgx \\ tgx \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ ctgx \leq 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ tgx \geq - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ ctgx \leq 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} tgx + ctgx = 2 \\ tgx + ctgx = 2 \\ tgx \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ ctgx \leq 2\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} tgx + ctgx = 2 \\ tgx \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ ctgx \leq 2\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[tgx + ctgx = 2\]

\[tgx + \frac{1}{\text{tgx}} = 2\]

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

\[(tgx - 1)^{2} = 0\]

\[tgx = 1\]

\[x = \frac{\pi}{4} + \pi k.\]

\[Ответ:\ x = \frac{\pi}{4} + \pi k.\]