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МатематикаРешебник по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 46
Задание 46
\[\boxed{\mathbf{46.}}\]
\[\textbf{а)}\ \sqrt[4]{x^{2} - 11x + 31} > \sqrt[4]{x - 4}\]
\[0 \leq x - 4 < x^{2} - 11x + 31\]
\[\left\{ \begin{matrix} x - 4 \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x - 4 < x^{2} - 11x + 31 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x \geq 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 12x + 35 > 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 12x + 35 > 0\]
\[D_{1} = 36 - 35 = 1\]
\[x_{1} = 6 + 1 = 7;\]
\[x_{2} = 6 - 1 = 5;\]
\[(x - 5)(x - 7) > 0\]
\[x < 5;\ \ \ x > 7.\]
\[x \in \lbrack 4;5) \cup (7; + \infty).\]
\[Ответ:\ x \in \lbrack 4;5) \cup (7; + \infty).\]
\[\textbf{б)}\ \sqrt[10]{x^{2} - 9} > \sqrt[10]{9x + 1}\]
\[0 \leq 9x + 1 < x^{2} - 9\]
\[\left\{ \begin{matrix} 9x + 1 \geq 0\ \ \ \ \ \ \ \ \ \ \\ x^{2} - 9 > 9x + 1 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x \geq - \frac{1}{9}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x^{2} - 9x - 10 > 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 9x - 10 > 0\]
\[x_{1} + x_{2} = 9;\ \ x_{1} \cdot x_{2} = - 10\]
\[x_{1} = 10;\ \ \ x_{2} = - 1.\]
\[(x + 1)(x - 10) > 0\]
\[x > 10.\]
\[Ответ:x > 10.\]
\[\textbf{в)}\ \sqrt[8]{x^{2} - 36} > \sqrt[8]{5x}\]
\[0 \leq 5x < x^{2} - 36\]
\[\left\{ \begin{matrix} 5x \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 36 > 5x \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 5x - 36 > 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 5x - 36 > 0\]
\[x_{1} + x_{2} = 5;x_{1} \cdot x_{2} = - 36\]
\[x_{1} = 9;\ \ x_{2} = - 4;\]
\[(x + 4)(x - 9) > 0\]
\[x < - 4;\ \ \ x > 9.\]
\[x > 9.\]
\[Ответ:x > 9.\]
\[\textbf{г)}\ \sqrt[4]{x + 19} > \sqrt[4]{49 - x^{2}}\]
\[0 \leq 49 - x^{2} < x + 19\]
\[\left\{ \begin{matrix} 49 - x^{2} \geq 0\ \ \ \ \ \ \ \ \ \ \ \\ 49 - x^{2} < x + 19 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} (x + 7)(x - 7) \leq 0 \\ x^{2} + x - 30 > 0\ \ \ \ \\ \end{matrix} \right.\ \]
\[- 7 \leq x \leq 7\]
\[x^{2} + x - 30 > 0\]
\[x_{1} + x_{2} = - 1;\ \ x_{1} \cdot x_{2} = - 30\]
\[x_{1} = - 6;\ \ x_{2} = 5;\]
\[(x + 6)(x - 5) > 0\]
\[x < - 6;\ \ x > 5.\]
\[x \in \lbrack - 7;\ - 6) \cup (5;7\rbrack.\]
\[Ответ:\ x \in \lbrack - 7;\ - 6) \cup (5;7\rbrack.\]
