Решебник по алгебре 11 класс Никольский Параграф 7. Равносильность уравнений и неравенств Задание 19

\[\boxed{\mathbf{19.}}\]

\[\textbf{а)}\ x^{3} - 5x^{2} + 4x > (x - 1)^{3}\]

\[x^{3} - 5x^{2} + 4x > x^{3} -\]

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

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

\[2x^{2} - x - 1 < 0\]

\[2x^{2} - x - 1 = 0\]

\[D = 1 + 8 = 9\]

\[x_{1} = \frac{1 + 3}{4} = 1;\]

\[x_{2} = \frac{1 - 3}{4} = - \frac{1}{2};\]

\[2 \cdot (x + 0,5)(x - 1) < 0\]

\[- 0,5 < x < 1.\]

\[\textbf{б)}\ x^{3} - 6x^{2} + 9x > (x - 3)^{3}\]

\[x^{3} - 6x^{2} + 9x > x^{3} - 9x^{2} +\]

\[+ 27x - 27\]

\[3x^{2} - 18x + 27 > 0\ \ \ |\ :3\]

\[x^{2} - 6x + 9 > 0\]

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

\[x \neq 3.\]

\[\textbf{в)}\ x^{3} + 5x^{2} - 6x - 2 < 3x^{2} -\]

\[- 3x + 4\]

\[x^{3} + 2x^{2} - 3x - 6 < 0\]

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

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

\[(x + 2)\left( x + \sqrt{3} \right)\left( x - \sqrt{3} \right) < 0\]

\[x < - 2;\]

\[- \sqrt{3} < x < \sqrt{3}.\]

\[\textbf{г)}\ 2x^{3} + 3x^{2} - 4x - 6 > x^{3} - 2x\]

\[x^{3} + 3x^{2} - 2x - 6 > 0\]

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

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

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

\[- 3 < x < - \sqrt{2};\]

\[x > \sqrt{2}.\]