Решебник по алгебре 11 класс Никольский Параграф 7. Равносильность уравнений и неравенств Задание 23

\[\boxed{\mathbf{23.}}\]

\[\textbf{а)}\ x + 1 > \sqrt[3]{x^{3} + 2x^{2} - 3x - 4}\]

\[(x + 1)^{3} > x^{3} + 2x^{2} - 3x - 4\]

\[x^{3} + 3x^{2} + 3x + 1 - x^{3} - 2x^{2} + 3x + 4 > 0\]

\[x^{2} + 6x + 5 > 0\]

\[x^{2} + 6x + 5 = (x + 5)(x + 1)\]

\[D_{1} = 9 - 5 = 4\]

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

\[x_{2} = - 3 - 2 = - 5.\]

\[(x + 5)(x + 1) > 0\]

\[x < - 5;\ \ x > - 1.\]

\[\textbf{б)}\ x + 2 < \sqrt[3]{x^{3} + 5x^{2} + 7x + 2}\]

\[(x + 2)^{3} < x^{3} + 5x^{2} + 7x + 2\]

\[x^{3} + 6x^{2} + 12x + 8 - x^{3} - 5x^{2} - 7x - 2 < 0\]

\[x^{2} + 5x + 6 < 0\]

\[x^{2} + 5x + 6 = (x + 3)(x + 2)\]

\[x_{1} + x_{2} = - 5;\ \ x_{1} \cdot x_{2} = 6\]

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

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

\[- 3 < x < - 2.\]