ГДЗ по алгебре 11 класс Никольский Параграф 7. Равносильность уравнений и неравенств Задание 8

\[\boxed{\mathbf{8.}}\]

\[\textbf{а)}\ \left( 5sin^{2}x - 4 \right)^{11} = \left( \text{si}n^{2}x - 1 \right)^{11}\]

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

\[4\sin^{2}x = 3\]

\[\sin^{2}x = \frac{3}{4}\]

\[\frac{1 - \cos{2x}}{2} = \frac{3}{4}\]

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

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

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

\[2x = \pm \arccos\left( - \frac{1}{2} \right) + 2\pi n\]

\[\arccos\left( - \frac{1}{2} \right) = \pi - \arccos\frac{1}{2} =\]

\[= \pi - \frac{\pi}{3} = \frac{2\pi}{3};\]

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

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

\[\textbf{б)}\ \left( 5\cos^{2}x - 1 \right)^{7} =\]

\[= \left( \cos^{2}x + 1 \right)^{7}\]

\[5\cos^{2}x - 1 = \cos^{2}x + 1\]

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

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

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

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

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

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

\[x = \frac{\pi}{4} + \frac{\text{πn}}{2}.\]

\[\textbf{в)}\ \left( 4^{x} - 5 \right)^{99} = \left( 3 \cdot 2^{x} - 1 \right)^{99}\]

\[4^{x} - 5 = 3 \cdot 2^{x} - 1\]

\[\left( 2^{2} \right)^{x} - 3 \cdot 2^{x} - 4 = 0\]

\[\left( 2^{x} \right)^{2} - 3 \cdot 2^{x} - 4 = 0\]

\[t = 2^{x}:\]

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

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

\[t_{1} = 4;\ \ \ t_{2} = - 1\]

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

\[нет\ корней.\]

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

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

\[x = 2.\]

\[Ответ:x = 2.\]

\[\textbf{г)}\ \left( 9^{x} - 1 \right)^{95} = \left( 3^{x} + 5 \right)^{95}\]

\[9^{x} - 1 = 3^{x} + 5\]

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

\[\left( 3^{x} \right)^{2} - 3^{x} - 6 = 0\]

\[t = 3^{x}:\]

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

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

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

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

\[нет\ корней.\]

\[3^{x} = 3\]

\[3^{x} = 3^{1}\]

\[x = 1.\]

\[Ответ:x = 1.\]