Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 14

\[\boxed{\mathbf{14.}}\]

\[\textbf{а)}\ \sqrt{6x - 3} < |x + 1|\]

\[6x - 3 \geq 0\]

\[6x \geq 3\]

\[x \geq 0,5.\]

\[M = \lbrack 0,5; + \infty).\]

\[6x - 3 < (x + 1)^{2}\]

\[6x - 3 < x^{2} + 2x + 1\]

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

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

\[x \neq 2.\]

\[Ответ:x \in \lbrack 0,5;2) \cup (2; + \infty).\]

\[\textbf{б)}\ \sqrt{6 + 3x} > |2x + 1|\]

\[6 + 3x \geq 0\]

\[3x \geq - 6\]

\[x \geq - 2.\]

\[M = \lbrack - 2; + \infty).\]

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

\[6 + 3x > 4x^{2} + 4x + 1\]

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

\[D = 1 + 80 = 81\]

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

\[x_{2} = \frac{- 1 - 9}{8} = - 1,25.\]

\[(x + 1,25)(x - 1) < 0\]

\[- 1,25 < x < 1.\]

\[Ответ:x \in \lbrack - 1,25;1).\]

\[\textbf{в)}\ \sqrt{5x - 1} > |3x - 1|\]

\[5x - 1 \geq 0\]

\[5x \geq 1\]

\[x \geq 0,2.\]

\[M = \lbrack 0,2; + \infty).\]

\[5x - 1 > (3x - 1)^{2}\]

\[5x - 1 > 9x^{2} - 6x + 1\]

\[9x^{2} - 11x + 2 < 0\]

\[D = 121 - 72 = 49\]

\[x_{1} = \frac{11 + 7}{18} = 1;\]

\[x_{2} = \frac{11 - 7}{18} = \frac{4}{18} = \frac{2}{9}.\]

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

\[\frac{2}{9} < x < 1.\]

\[Ответ:x \in \lbrack 0,2;1).\]

\[\textbf{г)}\ \sqrt{7x + 2} > |x + 2|\]

\[7x + 2 \geq 0\]

\[7x \geq - 2\]

\[x \geq - \frac{2}{7}.\]

\[M = \left\lbrack - \frac{2}{7}; + \infty \right).\]

\[7x + 2 > (x + 2)^{2}\]

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

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

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

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

\[(x - 1)(x - 2) < 0\]

\[1 < x < 2.\]

\[Ответ:x \in (1;2).\]