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

\[\boxed{\mathbf{11.}}\]

\[\textbf{а)}\ \sqrt{6x + 7} > x + 2\]

\[6x + 7 \geq 0\]

\[x \geq - \frac{7}{6}.\]

\[M = \left\lbrack - 1\frac{1}{6}; + \infty \right).\]

\[6x + 7 > (x + 2)^{2}\]

\[6x + 7 > x^{2} + 4x + 4\]

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

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

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

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

\[(x + 1)(x - 3) < 0\]

\[- 1 < x < 3.\]

\[x \in ( - 1;3).\]

\[Ответ:\ x \in ( - 1;3).\]

\[\textbf{б)}\ \sqrt{5x + 6} > x + 2\]

\[5x + 6 \geq 0\]

\[x \geq - \frac{6}{5}\]

\[x \geq - 1,2.\]

\[M = \lbrack - 1,2; + \infty).\]

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

\[5x + 6 > x^{2} + 4x + 4\]

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

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

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

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

\[- 1 < x < 2.\]

\[x \in ( - 1;2).\]

\[Ответ:\ x \in ( - 1;2).\]

\[\textbf{в)}\ \sqrt{6x - 2} > x + 1\]

\[6x - 2 \geq 0\]

\[x \geq \frac{1}{3}.\]

\[M = \left\lbrack \frac{1}{3}; + \infty \right).\]

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

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

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

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

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

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

\[(x - 1)(x - 3) < 0\]

\[1 < x < 3.\]

\[x \in (1;3).\]

\[Ответ:\ x \in (1;3).\]

\[\textbf{г)}\ \sqrt{5x - 1} > x + 1\]

\[5x - 1 \geq 0\]

\[5x \geq 1\]

\[x \geq 0,2.\]

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

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

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

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

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

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

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

\[1 < x < 2.\]

\[x \in (1;2).\]

\[Ответ:\ x \in (1;2)\]