Решебник по алгебре 11 класс Никольский Параграф 13. Использование свойств функции при решении уравнений и неравенств Задание 22

\[\boxed{\mathbf{22.}}\]

\[\textbf{а)}\ \frac{\sqrt{\sin^{2}x + 5} + \sqrt{\cos^{2}x + 4}}{2} =\]

\[= \sqrt[4]{\left( \sin^{2}x + 5 \right)\left( \cos^{2}x + 4 \right)}\]

\[\left( \sqrt[4]{\text{si}n^{2}x + 5} - \sqrt[4]{\text{co}s^{2}x + 4} \right)^{2} =\]

\[= 0\]

\[\sqrt[4]{\text{si}n^{2}x + 5} - \sqrt[4]{\text{co}s^{2}x + 4} = 0\]

\[\sqrt[4]{\text{si}n^{2}x + 5} = \sqrt[4]{\text{co}s^{2}x + 4}\]

\[\text{si}n^{2}x + 5 = cos^{2}x + 4\]

\[\text{co}s^{2}x - sin^{2}x = 1\]

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

\[2x = 2\pi n\]

\[x = \pi n.\]

\[Ответ:x = \pi n.\]

\[\sqrt[4]{\left( \sin^{2}x + \sin x + 5 \right)} =\]

\[= \sqrt[4]{\left( \sin x + 6 \right)}\]

\[\sin^{2}x + \sin x + 5 = \sin x + 6\]

\[\text{si}n^{2}x = 1\]

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

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

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

\[2x = \pi + 2\pi n\]

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

\[Ответ:x = \frac{\pi}{2} + \pi n.\]

\[\textbf{в)}\ \sqrt{3\log_{2}^{2}x + 5} + \sqrt{\log_{2}^{4}x + 1} =\]

\[= 2\sqrt[4]{\left( 3\log_{2}^{2}x + 5 \right)\left( \log_{2}^{4}x + 1 \right)}\]

\[\sqrt[4]{\left( 3\log_{2}^{2}x + 5 \right)} = \sqrt[4]{\left( \log_{2}^{4}x + 1 \right)}\]

\[3\log_{2}^{2}x + 5 = \log_{2}^{4}x + 1\]

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

\[\log_{2}^{2}x = y:\]

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

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

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

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

\[нет\ корней.\]

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

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

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

\[Ответ:\ \ x = \frac{1}{4};\ \ x = 4.\]

\[\textbf{г)}\ \sqrt{2 - \frac{1}{2}\lg^{2}x} + \sqrt{\lg^{4}x + \frac{1}{2}} =\]

\[= 2\sqrt[4]{\left( 2 - \frac{1}{2}\lg^{2}x \right)\left( \lg^{4}x + \frac{1}{2} \right)}\]

\[M = (0; + \infty).\]

\[\sqrt[4]{\left( 2 - \frac{1}{2}\lg^{2}x \right)} = \sqrt[4]{\left( \lg^{4}x + \frac{1}{2} \right)}\ \]

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

\[\lg^{4}x + \frac{1}{2}\lg^{2}x - \frac{3}{2} = 0\ \ | \cdot 2\]

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

\[\lg^{2}x = y:\]

\[D = 1 + 24 = 25\]

\[y_{1} = \frac{- 1 + 5}{4} = 1;\]

\[y_{2} = \frac{- 1 - 5}{4} =\]

\[= - 1,5\ (не\ подходит).\]

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

\[\lg x = \pm 1.\]

\[x = \frac{1}{10};\ \ x = 10.\]

\[Ответ:\ \ x = 0,1;x = 10.\]