Решебник по алгебре 11 класс Никольский Параграф 9. Равносильность уравнений и неравенств системам Задание 26

\[\boxed{\mathbf{26.}}\]

\[\textbf{а)}\log_{2}(x - 2) + \log_{2}(x - 3) =\]

\[= \log_{2}{(x^{2} - 5x + 6)}\]

\[\left\{ \begin{matrix} (x - 2)(x - 3) = x^{2} - 5x + 6 \\ (x - 2)(x - 3) > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x^{2} - 2x - 3x + 6 = x^{2} - 5x + 6 \\ x > 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 0x = 0 \\ x > 3\ \ \\ \end{matrix} \right.\ \]

\[Ответ:x > 3.\]

\[\textbf{б)}\log_{3}(x - 3) + \log_{3}(4 - x) =\]

\[= \log_{3}{( - x^{2} + 7x - 12)}\]

\[\left\{ \begin{matrix} (x - 3)(4 - x) = - x^{2} + 7x - 12 \\ (x - 3)(4 - x) > 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 4x - 12 - x^{2} + 3x = - x^{2} + 7x - 12 \\ 3 < x < 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 0x = 0\ \ \ \ \ \ \\ 3 < x < 4 \\ \end{matrix} \right.\ \]

\[Ответ:3 < x < 4.\]

\[x^{2} - 5x + 6 = 0\]

\[x_{1} + x_{2} = 5;\ \ x_{1} \cdot x_{2} = 6\]

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

\[(x - 2)(x - 3) > 0\]

\[x < 2;x > 3.\]

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

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

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

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

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

\[1 < x < 4.\]

\[\left( x^{2} - 5x + 6 \right)\left( 5x - x^{2} - 4 \right) =\]

\[= \left( x^{2} - 5x + 6 \right)\left( 5x - x^{2} - 4 \right)\]

\[имеет\ бесконечно\ много\]

\[\ решений\ (правая\ часть\ равна\ левой).\]

\[Ответ:1 < x < 2;3 < x < 4.\]