Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 15

\[\boxed{\mathbf{15.}}\]

\[\textbf{а)}\ \sqrt{x} < \sqrt[4]{x + 3}\]

\[x \geq 0;\]

\[x + 3 \geq 0\]

\[x \geq - 3.\]

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

\[\left( \sqrt{x} \right)^{4} < \left( \sqrt[4]{x + 3} \right)^{4}\]

\[x^{2} < x + 3\]

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

\[D = 1 + 12 = 13\]

\[x_{1} = \frac{1 + \sqrt{13}}{2};\]

\[x_{2} = \frac{1 - \sqrt{13}}{2};\]

\[\frac{1 - \sqrt{13}}{2} < x < \frac{1 + \sqrt{13}}{2}.\]

\[Ответ:x \in \left\lbrack 0;\ \frac{1 + \sqrt{13}}{2} \right).\]

\[\textbf{б)}\ \sqrt{x} > \sqrt[4]{x + 4}\]

\[x + 4 \geq 0\]

\[x \geq - 4.\]

\[x \geq 0.\]

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

\[\left( \sqrt{x} \right)^{4} > \left( \sqrt[4]{x + 4} \right)^{4}\]

\[x^{2} > x + 4\]

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

\[D = 1 + 16 = 17\]

\[x_{1} = \frac{1 + \sqrt{17}}{2};\]

\[x_{2} = \frac{1 - \sqrt{17}}{2};\]

\[x < \frac{1 - \sqrt{17}}{2};\]

\[x > \frac{1 + \sqrt{17}}{2}.\]

\[Ответ:x \in \left( \frac{1 + \sqrt{17}}{2}; + \infty \right).\]

\[\textbf{в)}\ \sqrt{x + 1} > \sqrt[3]{2x + 1}\]

\[x + 1 \geq 0\]

\[x \geq - 1.\]

\[M = \lbrack - 1; + \infty).\]

\[\left( \sqrt{x + 1} \right)^{6} > \left( \sqrt[3]{2x + 1} \right)^{6}\]

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

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

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

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

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

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

\[D = 1 + 4 = 5\]

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

\[x_{2} = \frac{1 - \sqrt{5}}{2};\]

\[x\left( x - \frac{1 - \sqrt{5}}{2} \right)\left( x - \frac{1 + \sqrt{5}}{2} \right) > 0\]

\[x \in \left( \frac{1 - \sqrt{5}}{2};0 \right) \cup \left( \frac{1 + \sqrt{5}}{2}; + \infty \right).\]

\[Ответ:\ x \in \lbrack - 1;0) \cup \left( \frac{1 + \sqrt{5}}{2}; + \infty \right).\]

\[\textbf{г)}\ \sqrt{x} < \sqrt[3]{3x - 2}\]

\[x \geq 0\]

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

\[\left( \sqrt{x} \right)^{6} < \left( \sqrt[3]{3x - 2} \right)^{6}\]

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

\[x^{3} < 9x^{2} - 12x + 4\]

\[x^{3} - 9x^{2} + 12x - 4 < 0\]

\[\left( x^{3} - 1 \right) - 3\left( 3x^{2} - 4x + 1 \right) < 0\]

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

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

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

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

\[(x - 1)\left( x^{2} + x + 1 - 9x + 3 \right) < 0\]

\[(x - 1)\left( x^{2} - 8x + 4 \right) < 0\]

\[x^{2} - 8x + 4 = 0\]

\[D_{1} = 16 - 4 = 12\]

\[x_{1} = 4 + \sqrt{12} = 4 + 2\sqrt{3};\]

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

\[x < 4 - 2\sqrt{3};\ 1 < x < 4 + 2\sqrt{3}.\]

\[Ответ:x \in \left( 1;4 + 2\sqrt{3} \right).\]