Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 8

\[\boxed{\mathbf{8.}}\]

\[\textbf{а)}\ \sqrt{x + 1} < x - 1\]

\[x - 1 \geq 0\]

\[x \geq 1.\]

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

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

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

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

\[x(x - 3) > 0\]

\[x < 0;x > 3.\]

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

\(Ответ:x \in (3; + \infty).\)

\[\textbf{б)}\ \sqrt{x + 4} < x + 2\]

\[x + 2 \geq 0\]

\[x \geq - 2.\]

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

\[x + 4 < (x + 2)^{2}\]

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

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

\[x(x + 3) > 0\]

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

\[x \in (0; + \infty).\]

\(Ответ:x \in (0; + \infty).\)

\[\textbf{в)}\ \sqrt{x + 1} < x + 1\]

\[x + 1 \geq 0\]

\[x \geq - 1.\]

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

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

\[x + 1 < x^{2} + 2x + 1\]

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

\[x(x + 1) > 0\]

\[x < - 1;\ \ x > 0.\]

\[x \in (0; + \infty).\]

\(Ответ:x \in (0; + \infty).\)

\[\textbf{г)}\ \sqrt{x + 4} < x - 2\]

\[x - 2 \geq 0\]

\[x \geq 2.\]

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

\[x + 4 < (x - 2)^{2}\]

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

\[x^{2} - 5x > 0\]

\[x(x - 5) > 0\]

\[x < 0;\ \ \ x > 5.\]

\[x \in (5; + \infty).\]

\(Ответ:x \in (5; + \infty)\)