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МатематикаРешебник по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 71
Задание 71
\[\boxed{\mathbf{71.}}\]
\[\left\{ \begin{matrix} \log_{2}(x - 3) < \log_{2}\left( x^{2} - 3x \right) \\ \log_{2}(x - 3) \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \log_{2}\left( x^{2} - 3x \right) \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \ \]
\[\left\{ \begin{matrix} x - 3 < x^{2} - 3x \\ x - 3 \geq 1\ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 3x \geq 1\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} - 4x + 3 > 0 \\ x \geq 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 3x - 1 \geq 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 4x + 3 > 0\]
\[D_{1} = 4 - 3 = 1\]
\[x_{1} = 2 + 1 = 3;\]
\[x_{2} = 2 - 1 = 1;\]
\[(x - 1)(x - 3) > 0\]
\[x > 3;\ \ x < 1.\]
\[x^{2} - 3x - 1 \geq 0\]
\[D = 9 + 4 = 13\]
\[x_{1} = \frac{3 + \sqrt{13}}{2};\]
\[x_{2} = \frac{3 - \sqrt{13}}{2};\]
\[\left\{ \begin{matrix} x > 3\ \ \ \ \ \ \ \ \ \ \ \ \\ x < 1\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x \geq 4\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x > \frac{3 + \sqrt{13}}{2} \\ x < \frac{3 - \sqrt{13}}{2} \\ \end{matrix} \right.\ \]
\[x \geq 4.\]
\[Ответ:x \geq 4.\]
\[\left\{ \begin{matrix} \log_{5}(x + 1) < \log_{5}\frac{2}{x} \\ \log_{5}(x + 1) \geq 0\ \ \ \ \ \ \ \ \\ \log_{5}\left( \frac{2}{x} \right) \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x + 1 < \frac{2}{x} \\ x + 1 \geq 1 \\ \frac{2}{x} \geq 1\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} \frac{x^{2} + x - 2}{x} < 0 \\ x \geq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{2 - x}{x} \geq 0\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x^{2} + x - 2 < 0 \\ x > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2 - x \geq 0\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[x^{2} + x - 2 < 0\]
\[x_{1} + x_{2} = - 1;\ \ x_{1} \cdot x_{2} = - 2\]
\[x_{1} = - 2;\ \ x_{2} = 1;\]
\[(x + 2)(x - 1) < 0\]
\[- 2 < x < 1.\]
\[\left\{ \begin{matrix} - 2 < x < 1 \\ x > 0\ \ \ \ \ \ \ \ \ \ \ \\ x \leq 2\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[0 < x < 1.\]
\[Ответ:x \in (0;1).\]
