ГДЗ по алгебре и начала математического анализа 10 класс Алимов Задание 167

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\[1)\ \sqrt{x - 2} > 3\]

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

\[x - 2 > 9\]

\[x > 11.\]

\[Ответ:\ \ x > 11.\]

\[2)\ \sqrt{x - 2} < 1\]

\[\left( \sqrt{x - 2} \right)^{2} < 1^{2}\]

\[x - 2 < 1\]

\[x < 3.\]

\[Выражение\ имеет\ смысл\ при:\]

\[x - 2 \geq 0\]

\[x \geq 2.\]

\[Ответ:\ \ 2 \leq x < 3.\]

\[3)\ \sqrt{3 - x} < 5\]

\[\left( \sqrt{3 - x} \right)^{2} < 5^{2}\]

\[3 - x < 25\]

\[- x < 22\]

\[x > - 22.\]

\[Выражение\ имеет\ смысл\ при:\]

\[3 - x \geq 0\]

\[x \leq 3.\]

\[Ответ:\ \ - 22 < x \leq 3.\]

\[4)\ \sqrt{4 - x} > 3\]

\[\left( \sqrt{4 - x} \right)^{2} > 3^{2}\]

\[4 - x > 9\]

\[- x > 5\]

\[x < - 5.\]

\[Ответ:\ \ x < - 5.\]

\[5)\ \sqrt{2x - 3} > 4\]

\[\left( \sqrt{2x - 3} \right)^{2} > 4^{2}\]

\[2x - 3 > 16\]

\[2x > 19\]

\[x > 9,5.\]

\[Ответ:\ \ x > 9,5.\]

\[6)\ \sqrt{x + 1} \geq \frac{2}{3}\]

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

\[x + 1 \geq \frac{4}{9}\]

\[9(x + 1) \geq 4\]

\[9x + 9 \geq 4\]

\[9x \geq - 5\]

\[x \geq - \frac{5}{9}.\]

\[Ответ:\ \ x \geq - \frac{5}{9}.\]

\[7)\ \sqrt{3x - 5} < 5\]

\[\left( \sqrt{3x - 5} \right)^{2} < 5^{2}\]

\[3x - 5 < 25\]

\[3x < 30\]

\[x < 10.\]

\[Выражение\ имеет\ смысл\ при:\]

\[3x - 5 \geq 0\]

\[3x \geq 5\]

\[x \geq \frac{5}{3}.\]

\[Ответ:\ \ 1\frac{2}{3} \leq x < 10.\]

\[8)\ \sqrt{4x + 5} \leq \frac{1}{2}\]

\[\left( \sqrt{4x + 5} \right)^{2} \leq \left( \frac{1}{2} \right)^{2}\]

\[4x + 5 \leq \frac{1}{4}\]

\[4(4x + 5) \leq 1\]

\[16x + 20 \leq 1\]

\[16x \leq - 19\]

\[x \leq - \frac{19}{16}.\]

\[Выражение\ имеет\ смысл\ при:\]

\[4x + 5 \geq 0\]

\[4x \geq - 5\]

\[x \geq - \frac{5}{4}.\]

\[Ответ:\ - 1\frac{1}{4} \leq x \leq - 1\frac{3}{16}.\]