ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 633

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\[1)\ \frac{\sqrt{1 - x^{3}} - 1}{1 + x} \leq x\]

\[ОДЗ:\]

\[\left\{ \begin{matrix} 1 - x^{3} \geq 0 \\ 1 + x \neq 0\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ }\left\{ \begin{matrix} x^{3} \leq 1 \\ x \neq - 1 \\ \end{matrix} \right.\ \]

\[x \leq 1;\ \ \ x \neq - 1.\]

\[\frac{\sqrt{1 - x^{3}} - 1}{1 + x} - x \leq 0\]

\[\frac{\sqrt{1 - x^{3}} - 1 - x - x^{2}}{1 + x} \leq 0\]

\[\left\lbrack \begin{matrix} \left\{ \begin{matrix} \sqrt{1 - x^{3}} - 1 - x - x^{2} \geq 0 \\ 1 + x < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \\ \left\{ \begin{matrix} \sqrt{1 - x^{3}} - 1 - x - x^{2} \leq 0 \\ 1 + x > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \\ \end{matrix} \right.\ \]

\[\sqrt{1 - x^{3}} - 1 - x - x^{2} = 0\]

\[\sqrt{1 - x^{3}} = x^{2} + x + 1\]

\[1 - x^{3} = \left( x^{2} + x + 1 \right)^{2}\]

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

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

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

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

\[1)\ \left\{ \begin{matrix} \left( x^{2} + x + 1 \right)\left( - x^{2} - 2x \right) \geq 0 \\ x < 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[x^{2} + x + 1 \geq 0 - при\ любом\ \text{x.}\]

\[\left\{ \begin{matrix} - x^{2} - 2x \geq 0 \\ x < - 1\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x(x + 2) \leq 0 \\ x < - 1\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow - 2 \leq x < - 1.\]

\[2)\ \left\{ \begin{matrix} \left( x^{2} + x + 1 \right)\left( - x^{2} - 2x \right) \leq 0 \\ x > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} - x^{2} - 2x \leq 0 \\ x > - 1\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x(x + 2) \geq 0 \\ x > - 1\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow x \geq 0.\]

\[С\ учетом,\ что\ x \leq 1:\]

\[- 2 \leq x < - 1;\ \ 0 \leq x \leq 1.\]

\[2)\ \frac{4x^{2} - 9}{\sqrt{3x^{2} - 3}} \leq \frac{2}{3}x + 1\]

\[ОДЗ:\]

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

\[3 \cdot \left( x^{2} - 1 \right) > 0\]

\[3 \cdot (x + 1)(x - 1) > 0\]

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

\[\frac{4x^{2} - 9}{\sqrt{3x^{2} - 3}} \leq \frac{2x + 3}{3}\text{\ \ \ \ }\]

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

\[3 \cdot \left( 4x^{2} - 9 \right) \leq\]

\[\leq (2x + 3)\sqrt{3x^{2} - 3}\]

\[3 \cdot (2x - 3)(2x + 3) -\]

\[- (2x + 3)\sqrt{3x^{2} - 3} \leq 0\]

\[(2x + 3)\left( 6x - 9 - \sqrt{3x^{2} - 3} \right) \leq\]

\[\leq 0\]

\[6x - 9 - \sqrt{3x^{2} - 3} = 0\]

\[\sqrt{3x^{2} - 3} = 6x - 9\]

\[3x^{2} - 3 = (6x - 9)^{2}\]

\[3x^{2} - 3 = 36x^{2} - 108x + 81\]

\[33x^{2} - 108x + 84 = 0\ \ \ \ |\ :3\]

\[11x^{2} - 36x + 28 = 0\]

\[D_{1} = 324 - 308 = 16\]

\[x_{1} = \frac{18 + 4}{11} = 2;\ \ \ \ \]

\[x_{2} = \frac{18 - 4}{11} = \frac{14}{11}.\]

\[(2x + 3)(x - 2)\left( x - \frac{14}{11} \right) \leq 0\]

\[x \leq - 1,5;\ \ \frac{14}{11} \leq x \leq 2.\]

\[С\ учетом,\ что\ x < - 1;x > 1:\ \ \]

\[x \leq - 1,5;\ \frac{14}{11} \leq x \leq 2.\]