Алгебра
Английский
Биология
География
Геометрия
История
Русский
Обществознание
Физика
Химия
Информатика
Литература
МатематикаРешебник по алгебре и начала математического анализа 10 класс Алимов Задание 1561
Задание 1561
\[\boxed{\mathbf{1561}\mathbf{.}}\]
\[1)\ 16^{\sin^{2}x} + 16^{\cos^{2}x} = 10\ \ \ \ \ | \bullet 16^{\cos^{2}x}\]
\[16^{\sin^{2}x + \cos^{2}x} + 16^{2\cos^{2}x} = 10 \bullet 16^{\cos^{2}x}\]
\[16^{2\cos^{2}x} - 10 \bullet 16^{\cos^{2}x} + 16^{1} = 0\]
\[y = 16^{\cos^{2}x}:\]
\[y^{2} - 10y + 16 = 0\]
\[D = 100 - 64 = 36\]
\[y_{1} = \frac{10 - 6}{2} = 2;\]
\[y_{2} = \frac{10 + 6}{2} = 8.\]
\[1)\ 16^{\cos^{2}x} = 2\]
\[2^{4\cos^{2}x} = 2^{1}\]
\[4\cos^{2}x = 1\]
\[\cos^{2}x = \frac{1}{4}\]
\[\cos x = \pm \frac{1}{2}\]
\[x = \pm \arccos\frac{1}{2} + \pi n\]
\[x = \pm \frac{\pi}{3} + \pi n.\]
\[2)\ 16^{\cos^{2}x} = 8\]
\[2^{4\cos^{2}x} = 2^{3}\]
\[4\cos^{2}x = 3\]
\[\cos^{2}x = \frac{3}{4}\]
\[\cos x = \pm \frac{\sqrt{3}}{2}\]
\[x = \pm \arccos\frac{\sqrt{3}}{2} + \pi n\]
\[x = \pm \frac{\pi}{6} + \pi n.\]
\[Ответ:\ \ \pm \frac{\pi}{3} + \pi n;\ \ \pm \frac{\pi}{6} + \pi n.\]
\[2)\ \left( \sqrt{3 + \sqrt{8}} \right)^{x} + \left( \sqrt{3 - \sqrt{8}} \right)^{x} = 34\]
\[\left( \sqrt{\left( 1 + \sqrt{2} \right)^{2}} \right)^{x} + \left( \sqrt{\left( 1 - \sqrt{2} \right)^{2}} \right)^{x} = 34\]
\[\left| 1 + \sqrt{2} \right|^{x} + \left| 1 - \sqrt{2} \right|^{x} = 34\]
\[\left( 1 + \sqrt{2} \right)^{2x} - 34\left( 1 + \sqrt{2} \right)^{x} + 1 = 0\]
\[y = \left( 1 + \sqrt{2} \right)^{x}:\]
\[y^{2} - 34y + 1 = 0\]
\[D = 1156 - 4 = 1152 = 576 \bullet 2\]
\[y = \frac{34 \pm \sqrt{1152}}{2} = \frac{34 \pm 24\sqrt{2}}{2} =\]
\[= 17 \pm 12\sqrt{2} =\]
\[= 9 \pm 12\sqrt{2} + 8 = \left( 3 \pm 2\sqrt{2} \right)^{2} =\]
\[= \left( 1 \pm 2\sqrt{2} + 2 \right)^{2} = \left( 1 \pm \sqrt{2} \right)^{4}.\]
\[1)\ \left( 1 + \sqrt{2} \right)^{x} = \left( 1 - \sqrt{2} \right)^{4} =\]
\[= \frac{\left( 1 - \sqrt{2} \right)^{4} \bullet \left( 1 + \sqrt{2} \right)^{4}}{\left( 1 + \sqrt{2} \right)^{4}} =\]
\[= \frac{(1 - 2)^{4}}{\left( 1 + \sqrt{2} \right)^{4}} = \frac{1}{\left( 1 + \sqrt{2} \right)^{4}} =\]
\[= \left( 1 + \sqrt{2} \right)^{- 4}\]
\[x = - 4.\]
\[2)\ \left( 1 + \sqrt{2} \right)^{x} = \left( 1 + \sqrt{2} \right)^{4}\]
\[x = 4.\]
\[Ответ:\ \ x = \pm 4.\]
