Задача 26. Решите уравнение \({\log _9}{\left( {{x^2}-5x + 6} \right)^2} = \dfrac{1}{2}{\log _{\sqrt 3 }}\dfrac{{x-1}}{2} + {\log _3}\left| {\,x-3\,} \right|\)
ОТВЕТ: \(\dfrac{5}{3}.\)
\({\log _9}{\left( {{x^2}-5x + 6} \right)^2} = \dfrac{1}{2}{\log _{\sqrt 3 }}\dfrac{{x-1}}{2} + {\log _3}\left| {x-3} \right|.\) Запишем ОДЗ: \(\left\{ {\begin{array}{*{20}{c}}{{{\left( {{x^2}-5x + 6} \right)}^2} > 0,}\\{x-1 > 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left| {x-3} \right| > 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{x^2}-5x + 6 \ne 0,}\\{x > 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x-3 \ne 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\, \Leftrightarrow } \right.} \right.\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \ne 2,}\\{x \ne 3,}\\{x > 1}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x\, \in \,\left( {1;2} \right) \cup \left( {2;3} \right) \cup \left( {3;\infty } \right).} \right.\) \({\log _3}\left| {{x^2}-5x + 6} \right| = {\log _3}\dfrac{{x-1}}{2} + {\log _3}\left| {x-3} \right|\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\log _3}\left( {\left| {x-2} \right| \cdot \left| {x-3} \right|} \right) = {\log _3}\dfrac{{\left( {x-1} \right) \cdot \left| {x-3} \right|}}{2}\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,\left| {x-2} \right| \cdot \left| {x-3} \right|-\dfrac{{\left( {x-1} \right) \cdot \left| {x-3} \right|}}{2} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left| {x-3} \right| \cdot \left( {\left| {x-2} \right|-\dfrac{{x-1}}{2}} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left| {x-3} \right| = 0,\,\,\,\,\,\,\,\,\,\,}\\{2\left| {x-2} \right| = x-1}\end{array}} \right.\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x-3 = 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left\{ {\begin{array}{*{20}{c}}{x-1 \ge 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left[ {\begin{array}{*{20}{c}}{2x-4 = x-1,}\\{2x-4 = -x + 1}\end{array}} \right.}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 3,}\\{\left\{ {\begin{array}{*{20}{c}}{x \ge 1,\,\,\,\,}\\{\left[ {\begin{array}{*{20}{c}}{x = 3,}\\{x = \dfrac{5}{3}}\end{array}\,\,\,} \right.}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 3,}\\{x = \dfrac{5}{3}.}\end{array}} \right.\) Корень \(x = 3\) не удовлетворяет ОДЗ. Ответ: \(\dfrac{5}{3}\).