Решебник по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 47

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\[\textbf{а)}\ \sqrt[8]{x^{2} + \sqrt{x}} \geq \sqrt[8]{5x - 4 + \sqrt{x}}\]

\[0 \leq 5x - 4 + \sqrt{x} \leq x^{2} + \sqrt{x}\]

\[\left\{ \begin{matrix} 5x - 4 + \sqrt{x} \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + \sqrt{x} \geq 5x - 4 + \sqrt{x} \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 5x - 4 + \sqrt{x} \geq 0 \\ x^{2} - 5x + 4 \geq 0 \\ \end{matrix} \right.\ \]

\[5x - 4 + \sqrt{x} \geq 0\]

\[\sqrt{x} = t:\]

\[5t^{2} + t - 4 \geq 0\]

\[D = 1 + 80 = 81\]

\[t_{1} = \frac{- 1 + 9}{10} = \frac{8}{10} = \frac{4}{5};\]

\[t_{1} = \frac{- 1 - 9}{10} =\]

\[= - 1\ (не\ подходит).\]

\[\sqrt{x} \geq \frac{4}{5};\]

\[x \geq \frac{16}{25}.\]

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

\[x_{1} + x_{2} = 5;\ \ x_{1} \cdot x_{2} = 4\]

\[x_{1} = 1;\ \ x_{2} = 4.\]

\[(x - 1)(x - 4) \geq 0\]

\[x \leq 1;\ \ x \geq 4.\]

\[x \in \left\lbrack \frac{16}{25};1 \right\rbrack \cup \lbrack 4; + \infty).\]

\[Ответ:\ x \in \left\lbrack \frac{16}{25};1 \right\rbrack \cup \lbrack 4; + \infty).\]

\[\textbf{б)}\ \sqrt[6]{x^{2} + \sqrt{x}} \geq \sqrt[6]{3x^{2} - 2 + \sqrt{x}}\]

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

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

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

\[3x - 2 + \sqrt{x} \geq 0\]

\[\sqrt{x} = t \geq 0:\]

\[3t^{2} + x - 2 \geq 0\]

\[D = 1 + 24 = 25\]

\[t_{1} = \frac{- 1 + 5}{6} = \frac{2}{3};\]

\[t_{2} = \frac{- 1 - 5}{6} =\]

\[= - 1\ (не\ подходит).\]

\[\sqrt{x} \geq \frac{2}{3}\]

\[x \geq \frac{4}{9}.\]

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

\[x_{1} + x_{2} = 3;\ \ x_{1} \cdot x_{2} = 2\]

\[x_{1} = 1;\ \ x_{2} = 2.\]

\[(x - 1)(x - 2) \geq 0\]

\[x \leq 1;\ \ \ x \geq 2.\]

\[x \in \left\lbrack \frac{4}{9};1 \right\rbrack \cup \lbrack 2; + \infty).\]

\[Ответ:\ x \in \left\lbrack \frac{4}{9};1 \right\rbrack \cup \lbrack 2; + \infty).\]

\[\textbf{в)}\ \sqrt[10]{x^{2} - \sqrt{x}} \geq \sqrt[10]{5x - 4 - \sqrt{x}}\]

\[0 \leq 5x - 4 - \sqrt{x} \leq x^{2} - \sqrt{x}\]

\[\left\{ \begin{matrix} 5x - 4 - \sqrt{x} \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - \sqrt{x} \geq 5x - 4 - \sqrt{x} \\ \end{matrix} \right.\ \]

\[5x - 4 - \sqrt{x} \geq 0\]

\[\sqrt{x} = t \geq 0:\]

\[5t^{2} - x - 4 \geq 0\]

\[D = 1 + 80 = 81\]

\[t_{1} = \frac{1 + 9}{10} = 1;\]

\[t_{2} = \frac{1 - 9}{10} = - \frac{4}{5} < 0.\]

\[\sqrt{x} \geq 1.\]

\[\left\{ \begin{matrix} x \geq 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 5x + 4 \geq 0 \\ \end{matrix} \right.\ \]

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

\[x_{1} + x_{2} = 5;\ \ \ x_{1} \cdot x_{2} = 4\]

\[x_{1} = 1;\ \ \ x_{2} = 4.\]

\[(x - 1)(x - 4) \geq 0\]

\[x \leq 1;\ \ x \geq 4.\]

\[x \in \lbrack 1\rbrack \cup \lbrack 4; + \infty).\]

\[Ответ:\ x \in \lbrack 1\rbrack \cup \lbrack 4; + \infty).\]

\[\textbf{г)}\ \sqrt[12]{x^{2} - \sqrt{x}} \geq \sqrt[12]{3x - 2 - \sqrt{x}}\]

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

\[\left\{ \begin{matrix} 3x - 2 - \sqrt{x} \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - \sqrt{x} \geq 3x - 2 - \sqrt{x} \\ \end{matrix} \right.\ \]

\[3x - 2 - \sqrt{x} \geq 0\]

\[\sqrt{x} = t \geq 0:\]

\[3t^{2} - x - 2 \geq 0\]

\[D = 1 + 24 = 25\]

\[t_{1} = \frac{1 + 5}{6} = 1;\]

\[t_{2} = \frac{1 - 5}{6} = - \frac{2}{3} < 0.\]

\[\sqrt{x} \geq 1.\]

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

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

\[x_{1} + x_{2} = 3;\ \ x_{1} \cdot x_{2} = 2\]

\[x_{1} = 1;\ \ x_{2} = 2.\]

\[(x - 1)(x - 2) \geq 0\]

\[x \leq 1;\ \ x \geq 2.\]

\[x \in \lbrack 1\rbrack \cup \lbrack 2; + \infty).\]

\[Ответ:\ x \in \lbrack 1\rbrack \cup \lbrack 2; + \infty).\]