Решебник по алгебре 11 класс Никольский Параграф 7. Равносильность уравнений и неравенств Задание 12

\[\boxed{\mathbf{12.}}\]

\[\textbf{а)}\ 5^{x + \sin x} = 5^{\sin x + 2}\]

\[x + \sin x = \sin x + 2\]

\[x = 2.\]

\[\textbf{б)}\ 6^{\sqrt[3]{x + 1}} = 6^{\sqrt[3]{2x - 1}}\]

\[\sqrt[3]{x + 1} = \sqrt[3]{2x - 1}\]

\[x + 1 = 2x - 1\]

\[x = 2.\]

\[\textbf{в)}\ \left( x^{2} - \sin x \right)^{101} = \left( x^{2} + 1 \right)^{101}\]

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

\[- \sin x = 1\]

\[\sin x = - 1\]

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

\[\textbf{г)}\ \left( x^{7} + \cos x \right)^{103} = \left( x^{7} - 1 \right)^{103}\]

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

\[\cos x = - 1\]

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

\[\textbf{д)}\ \sqrt[7]{\sin^{2}x + 4^{x} - 6} =\]

\[= \sqrt[7]{\sin^{2}x - 2^{x}}\]

\[\sin^{2}x + 4^{x} - 6 = \sin^{2}x - 2^{x}\]

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

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

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

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

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

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

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

\[нет\ корней.\]

\[2^{x} = 2\]

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

\[x = 1.\]

\[\textbf{е)}\ \sqrt[5]{\sin^{2}x + 9^{x}} =\]

\[= \sqrt[5]{- cos^{2}x + 3^{x} + 7}\]

\[\sin^{2}x + 9^{x} = - cos^{2}x + 3^{x} + 7\]

\[\sin^{2}x + \cos^{2}x +\]

\[+ \left( 3^{2} \right)^{x} - 3^{x} - 7 = 0\]

\[1 + \left( 3^{x} \right)^{2} - 3^{x} - 7 = 0\]

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

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

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

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

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

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

\[нет\ корней.\]

\[3^{x} = 3\]

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

\[x = 1.\]