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МатематикаГДЗ по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 63
Задание 63
\[\boxed{\mathbf{63.}}\]
\[1)\ \left\{ \begin{matrix} 0 < x^{2} - 10x + 21 \leq 5 \\ 2|x| > 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} - 10x + 21 \leq 5 \\ x^{2} - 10x + 21 > 0 \\ |x| > \frac{1}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} - 10x + 16 \leq 0 \\ x^{2} - 10x + 21 > 0 \\ x > \frac{1}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x < - \frac{1}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]
\[x^{2} - 10x + 16 \leq 0\]
\[D_{1} = 25 - 16 = 9\]
\[x_{1} = 5 + 3 = 8;\]
\[x_{2} = 5 - 3 = 2.\]
\[(x - 2)(x - 8) \leq 0\]
\[2 \leq x \leq 8.\]
\[x^{2} - 10x + 21 > 0\]
\[D_{1} = 25 - 21 = 4\]
\[x_{1} = 5 + 2 = 7;\]
\[x_{2} = 5 - 2 = 3.\]
\[(x - 3)(x - 7) > 0\]
\[x < 3;\ \ x > 7.\]
\[x \in \lbrack 2;3) \cup (7;8\rbrack.\]
\[2)\ \left\{ \begin{matrix} x^{2} - 10x + 21 \geq 5 \\ 0 < 2|x| < 1\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} - 10x + 16 \geq 0 \\ 0 < |x| < \frac{1}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x \leq 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \geq 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - \frac{1}{2} < x < \frac{1}{2};x \neq 0 \\ \end{matrix} \right.\ \]
\[x \in \left( - \frac{1}{2};0 \right) \cup \left( 0;\frac{1}{2} \right).\]
\[Объединим\ полученные\ \]
\[результаты:\]
\[1)\ \left\{ \begin{matrix} 0 < x^{2} - 13x + 36 \leq 6 \\ 2|x| > 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} - 13x + 36 \leq 6 \\ x^{2} - 13x + 36 > 0 \\ |x| > \frac{1}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} - 13x + 30 \leq 0 \\ x^{2} - 13x + 36 > 0\ \\ x > \frac{1}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x < - \frac{1}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]
\[x^{2} - 13x + 30 \leq 0\]
\[x_{1} + x_{2} = 13;\ \ x_{1} \cdot x_{2} = 30\]
\[x_{1} = 3;\ \ x_{2} = 10.\]
\[(x - 3)(x - 10) \leq 0\]
\[3 \leq x \leq 10.\]
\[x^{2} - 13x + 36 > 0\]
\[x_{1} + x_{2} = 13;\ \ x_{1} \cdot x_{2} = 36\]
\[x_{1} = 9;\ \ \ x_{2} = 4.\]
\[(x - 4)(x - 9) > 0\]
\[x < - 4;\ \ x > 9.\]
\[x \in \lbrack 3;4) \cup (9;10\rbrack.\]
\[2)\ \left\{ \begin{matrix} x^{2} - 13x + 36 \geq 6 \\ 0 < 2|x| < 1\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} - 13x + 30 \geq 0 \\ 0 < |x| < \frac{1}{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x \leq 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \geq 10\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - \frac{1}{2} < x < \frac{1}{2};x \neq 0 \\ \end{matrix} \right.\ \]
\[x \in \left( - \frac{1}{2};0 \right) \cup \left( 0;\frac{1}{2} \right).\]
\[Объединим\ полученные\ \]
\[результаты:\]
\[\log_{|x|}{(x^{2} + 25)} \leq \log_{|x|}{226x^{2}}\]
\[1)\ \left\{ \begin{matrix} 0 < x^{2} + 25 \leq 226x^{2} \\ |x| > 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 25 \leq 226x^{2} - x^{2} \\ x^{2} + 25 > 0\ \ \ \ \ \ \ \ \ \\ x > 1;\ \ \ \ \ x < - 1 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} \geq \frac{1}{9}\text{\ \ \ \ \ \ } \\ x^{2} > - 25 \\ x > 1\ \ \ \ \ \ \ \\ x < - 1\ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x \geq \frac{1}{3}\text{\ \ \ \ \ } \\ x \leq - \frac{1}{3} \\ x > 1\ \ \ \ \\ x < - 1\ \\ \end{matrix} \right.\ \ \]
\[x \in ( - \infty; - 1) \cup (1; + \infty).\]
\[2)\ \left\{ \begin{matrix} x^{2} + 25 \geq 226x^{2} \\ 0 < |x| < 1\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 25 \geq 225x^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ } \\ - 1 < x < 1;x \neq 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} \leq \frac{1}{9}\text{\ \ \ \ \ \ \ \ \ } \\ - 1 < x < 1 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} - \frac{1}{3} \leq x \leq \frac{1}{3} \\ - 1 < x < 1 \\ \end{matrix} \right.\ ;\ \ x \neq 0\]
\[x \in \left\lbrack - \frac{1}{3};0 \right) \cup \left( 0;\frac{1}{3} \right\rbrack.\]
\[Объединим\ полученные\ \]
\[результаты:\]
\[\log_{|x|}\left( x^{2} + 361 \right) \leq \log_{|x|}{362x^{2}}\]
\[1)\ \left\{ \begin{matrix} 0 < x^{2} + 361 \leq 362x^{2} \\ |x| > 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 361 \leq 361x^{2} \\ x^{2} + 361 > 0 \\ |x| > 1\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} \geq 1\ \ \ \ \ \ \ \ \\ x^{2} > - 361 \\ x > 1\ \ \ \ \ \ \ \ \ \ \\ x < - 1\ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x \geq 1\ \ \ \\ x \leq - 1 \\ x > 1\ \ \ \\ x < - 1 \\ \end{matrix} \right.\ \ \]
\[x \in ( - \infty; - 1) \cup (1; + \infty).\]
\[2)\ \left\{ \begin{matrix} x^{2} + 361 \geq 362x^{2} \\ 0 < |x| < 1\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} 361x^{2} \leq 361 \\ 0 < |x| < 1\ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} \leq 1\ \ \ \ \ \ \ \ \ \\ - 1 < x < 1 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} - 1 \leq x \leq 1 \\ - 1 < x < 1 \\ \end{matrix} \right.\ ;\ \ x \neq 0\]
\[x \in ( - 1;0) \cup (0;1).\]
\[Объединим\ полученные\ \]
\[результаты:\]
