Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 30

\[\boxed{\mathbf{30.}}\]

\[\textbf{а)}\ \left( 2\sqrt{x} + 1 \right)^{2} > 5x^{2} + 4\sqrt{x} - 63\]

\[x \geq 0\]

\[M = \lbrack 0; + \infty).\]

\[4x + 4\sqrt{x} + 1 > 5x^{2} + 4\sqrt{x} - 63\]

\[5x^{2} - 4x - 64 < 0\]

\[D_{1} = 4 + 320 = 324 = 18^{2}\]

\[x_{1} = \frac{2 + 18}{5} = \frac{20}{5} = 4;\]

\[x_{2} = \frac{2 - 18}{5} = - \frac{16}{5} = - 3,2;\]

\[(x + 3,2)(x - 4) < 0\]

\[- 3,2 < x < 4.\]

\[Решение\ неравенства:\]

\[x \in \lbrack 0;4).\]

\[Ответ:\ x \in \lbrack 0;4).\]

\[\textbf{б)}\ \left( 2\sqrt{x} - 1 \right)^{2} < 2x^{2} - 4\sqrt{x} - 125\]

\[x \geq 0;\]

\[M = \lbrack 0; + \infty).\]

\[4x - 4\sqrt{x} + 1 < 2x^{2} - 4\sqrt{x} - 125\]

\[2x^{2} - 4x - 126 > 0\ \ \ |\ :2\]

\[x^{2} - 2x - 63 > 0\]

\[D_{1} = 1 + 63 = 64\]

\[x_{1} = 1 + 8 = 9;\]

\[x_{2} = 1 - 8 = - 7;\]

\[(x + 7)(x - 9) > 0\]

\[x < - 7;\ \ x > 9.\]

\[Решение\ неравенства:\]

\[x \in (9; + \infty).\]

\[Ответ:\ x \in (9; + \infty).\]

\[\textbf{в)}\ \left( 3\sqrt{x} + 2 \right)^{2} > 6x^{2} + 12\sqrt{x} - 2\]

\[x \geq 0;\]

\[M = \lbrack 0; + \infty).\]

\[9x + 12\sqrt{x} + 4 < 6x^{2} + 12\sqrt{x} - 2\]

\[6x^{2} - 9x - 6 > 0\ \ \ |\ :3\]

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

\[D = 9 + 16 = 25\]

\[x_{1} = \frac{3 + 5}{4} = 2;\]

\[x_{2} = \frac{3 - 5}{4} = - 0,5;\]

\[(x + 0,5)(x - 2) > 0\]

\[- 0,5 < x < 2.\]

\[Решение\ неравенства:\]

\[x \in \lbrack 0;2)\]

\[Ответ:\ x \in \lbrack 0;2).\]

\[\textbf{г)}\ \left( 3\sqrt{x} - 2 \right)^{2} < 4x^{2} - 12\sqrt{x} - 5\]

\[x \geq 0;\]

\[M = \lbrack 0; + \infty).\]

\[9x - 12\sqrt{x} + 4 < 4x^{2} - 12\sqrt{x} - 5\]

\[4x^{2} - 9x - 9 > 0\]

\[D = 81 + 144 = 225\]

\[x_{1} = \frac{9 + 15}{8} = 3;\]

\[x_{2} = \frac{9 - 15}{8} = - \frac{6}{8} = - \frac{3}{4} =\]

\[= - 0,75;\]

\[(x + 0,75)(x - 3) > 0\]

\[x < - 0,75;\ \ x > 3.\]

\[Решение\ неравенства:\]

\[x \in (3; + \infty).\]

\[Ответ:\ x \in (3; + \infty).\]