Решебник по алгебре 11 класс Никольский Параграф 11. Равносильность неравенств на множествах Задание 9

\[\boxed{\mathbf{9.}}\]

\[\textbf{а)}\ \sqrt{x + 1} > x - 1\]

\[x + 1 \geq 0\]

\[x \geq - 1.\]

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

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

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

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

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

\[0 < x < 3.\]

\[x \in \lbrack - 1;3)\]

\(Ответ:x \in \lbrack - 1;3).\)

\[\textbf{б)}\ \sqrt{x + 4} > x - 2\]

\[x + 4 \geq 0\]

\[x \geq - 4.\]

\[M = \lbrack - 4; + \infty).\]

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

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

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

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

\[0 < x < 5.\]

\[x \in \lbrack - 4;5).\]

\(Ответ:x \in \lbrack - 4;5)\text{.\ }\)

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

\[2x + 1 \geq 0\]

\[x \geq - 0,5.\]

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

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

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

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

\[x(x - 4) < 0\]

\[x < 0;\ \ x > 4.\]

\[x \in \lbrack - 0,5;4).\]

\(Ответ:x \in \lbrack - 0,5;4).\)

\[\textbf{г)}\ \sqrt{3x + 4} > x - 2\]

\[3x + 4 \geq 0\]

\[3x \geq - 4\]

\[x \geq - \frac{4}{3}.\]

\[M = \left\lbrack - \frac{4}{3}; + \infty \right).\]

\[3x + 4 > (x - 2)^{2}\]

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

\[x^{2} - 7x < 0\]

\[x(x - 7) < 0\]

\[x < 0;x > 7.\]

\[x \in \left\lbrack - \frac{4}{3};7 \right).\]

\[Ответ:\ x \in \left\lbrack - \frac{4}{3};7 \right).\]