Решебник по алгебре и начала математического анализа 11 класс Колягин Задание 927

\[1)\ \sqrt{9x - 20} < x\]

\[9x - 20 < x^{2}\]

\[x^{2} - 9x + 20 > 0\]

\[D = 81 - 80 = 1\]

\[x_{1} = \frac{9 - 1}{2} = 4;\]

\[x_{2} = \frac{9 + 1}{2} = 5;\]

\[(x - 4)(x - 5) > 0\]

\[x < 4;\ \ \ x > 5.\]

\[Область\ определения:\]

\[9x - 20 \geq 0\]

\[x \geq \frac{20}{9}.\]

\[Ответ:\ \ x \in \left\lbrack \frac{20}{9};\ 4 \right) \cup (5;\ + \infty).\]

\[2)\ \sqrt{x + 7} > x + 1\]

\[x + 1 < 0\]

\[x < - 1.\]

\[x + 7 > x^{2} + 2x + 1\ \ \ \]

\[x^{2} + x - 6 < 0\ \ \ \]

\[D = 1 + 24 = 25\]

\[x_{1} = \frac{- 1 - 5}{2} = - 3;\]

\[x_{2} = \frac{- 1 + 5}{2} = 2;\]

\[(x + 3)(x - 2) < 0\]

\[- 3 < x < 2.\]

\[Область\ определения:\]

\[x + 7 \geq 0\]

\[x \geq - 7.\]

\[Ответ:\ \ x \in \lbrack - 7;\ 2).\]

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

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

\[\frac{x^{2}(2 - x) - (x + 4)}{2 - x} < 0\]

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

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

\[\frac{(x + 1)\left( x^{2} - 3x + 4 \right)}{x - 2} < 0\]

\[D = 9 - 16 = - 7 < 0\]

\[x \in R.\]

\[\frac{x + 1}{x - 2} < 0\]

\[- 1 < x < 2.\]

\[Область\ определения:\]

\[\frac{x + 4}{2 - x} \geq 0\]

\[\frac{x + 4}{x - 2} \leq 0\]

\[- 4 \leq x < 2.\]

\[Ответ:\ \ x \in \lbrack - 4;\ 2).\]

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

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

\[\frac{\left( 1 - 2x + x^{2} \right)(1 + 2x) - (1 + 5x)}{1 + 2x} \geq 0\]

\[\frac{1 + 2x - 2x - 4x^{2} + x^{2} + 2x^{3} - 1 - 5x}{1 + 2x} \geq 0\]

\[\frac{2x^{3} - 3x^{2} - 5x}{1 + 2x} \geq 0\]

\[\frac{x\left( 2x^{2} - 3x - 5 \right)}{1 + 2x} \geq 0\]

\[D = 9 + 40 = 49\]

\[x_{1} = \frac{3 - 7}{2 \bullet 2} = - 1;\]

\[x_{2} = \frac{3 + 7}{2 \bullet 2} = 2,5;\]

\[\frac{(x + 1)x(x - 2,5)}{1 + 2x} \geq 0\]

\[x \leq - 1;\]

\[- 0,5 < x \leq 0;\]

\[x \geq 2,5.\]

\[Область\ определения:\]

\[\frac{1 + 5x}{1 + 2x} \geq 0\]

\[x < - \frac{1}{2};\ \ x \geq - \frac{1}{5};\ \ \ x \leq 1.\]

\[Ответ:\ \ x \in ( - \infty;\ - 1\rbrack \cup \left\lbrack - \frac{1}{5};\ 0 \right\rbrack.\]