Решебник по алгебре 11 класс Никольский Параграф 7. Равносильность уравнений и неравенств Задание 26

\[\boxed{\mathbf{26.}}\]

\[\textbf{а)}\ 2^{x + 1} > 2^{x^{2} - 5}\]

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

\[x^{2} - x - 6 < 0\]

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

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

\[(x + 2)(x - 3) < 0\]

\[- 2 < x < 3.\]

\[\textbf{б)}\ (0,3)^{2x + 5} > (0,3)^{x^{2} + 2}\]

\[2x + 5 < x^{2} + 2\]

\[x^{2} - 2x - 3 > 0\]

\[D_{1} = 1 + 3 = 4\]

\[x_{1} = 1 + 2 = 3;\]

\[x_{2} = 1 - 2 = - 1.\]

\[(x + 1)(x - 3) > 0\]

\[x < - 1;\ \ x > 3.\]

\[\textbf{в)}\ 5^{2x - 9} < 5^{x^{2} - 12}\]

\[2x - 9 < x^{2} - 12\]

\[x^{2} - 2x - 3 > 0\]

\[D_{1} = 1 + 3 = 4\]

\[x_{1} = 1 + 2 = 3;\]

\[x_{2} = 1 - 2 = - 1.\]

\[(x + 1)(x - 3) > 0\]

\[x < - 1;\ \ x > 3.\]

\[\textbf{г)}\ (0,5)^{4x - 7} < (0,5)^{x^{2} - 4}\]

\[4x - 7 > x^{2} - 4\]

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

\[D_{1} = 4 - 3 = 1\]

\[x_{1} = 2 + 1 = 3;\]

\[x_{2} = 2 - 1 = 1.\]

\[(x - 1)(x - 3) < 0\]

\[1 < x < 3.\]