ГДЗ по алгебре 11 класс Никольский Параграф 7. Равносильность уравнений и неравенств Задание 31

\[\boxed{\mathbf{31.}}\]

\[\textbf{а)}\ 4x + \sin x > \sin x + 2\]

\[4x > 2\]

\[x > \frac{1}{2}.\]

\[\textbf{б)}\ \sqrt[3]{3x - 50} < \sqrt[3]{2x + 20}\]

\[3x - 50 < 2x + 20\]

\[x < 70.\]

\[\textbf{в)}\ \sqrt[17]{\left( x^{2} - \sin x \right)^{17}} > \sqrt[17]{\left( x^{2} + 0,5 \right)^{17}}\]

\[x^{2} - \sin x > x^{2} + 0,5\]

\[- \sin x > 0,5\]

\[\sin x < - \frac{1}{2}\]

\[\textbf{г)}\ \sqrt[13]{\left( x^{8} + \cos x \right)^{13}} < \sqrt[13]{\left( x^{8} - 0,5 \right)^{13}}\]

\[{x^{8} + \cos x < x^{8} - 0,5 }{\cos x < - \frac{1}{2}}\]

\[\textbf{д)}\ \sqrt[3]{3\sin^{2}x + 4^{x} - 3} > \sqrt[3]{- 3\cos^{2}x + 2^{x}}\]

\[3\sin^{2}x + 4^{x} - 3 > - 3\cos^{2}x + 2^{x}\]

\[3\sin^{2}x + 4^{x} - 3 + 3\cos^{2}x - 2^{x} > 0\]

\[3 \cdot \left( \sin^{2}x + \cos^{2}x \right) + 4^{x} - 3 - 2^{x} > 0\]

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

\[\left( 2^{x} \right)^{2} - 2^{x} > 0\]

\[2^{x}\left( 2^{x} - 1 \right) > 0\]

\[2^{x} > 1\]

\[2^{x} > 2^{0}\]

\[x > 0.\]

\[\textbf{е)}\ \sqrt[5]{\sin^{2}x + 12 \cdot 3^{x} - 28} < \sqrt[5]{- \cos^{2}x + 9^{x}}\]

\[\sin^{2}x + 12 \cdot 3^{x} - 28 + \cos^{2}x - 9^{x} < 0\]

\[\left( \text{si}n^{2}x + cos^{2}x \right) + 12 \cdot 3^{x} - \left( 3^{2} \right)^{x} - 28 < 0\]

\[1 + 12 \cdot 3^{x} - \left( 3^{x} \right)^{2} - 28 < 0\]

\[\left( 3^{x} \right)^{2} - 12 \cdot 3^{x} + 27 > 0\]

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

\[t^{2} - 12t + 27 = 0\]

\[D_{1} = 36 - 27 = 9\]

\[t_{1} = 6 + 3 = 9;\]

\[t_{2} = 6 - 3 = 3.\]

\[t < 3:\]

\[3^{x} < 3\]

\[x < 1.\]

\[t > 9:\]

\[3^{x} > 9\]

\[3^{x} > 3^{2}\]

\[x > 2.\]