Задача 10. Решите уравнение \({\log _{\sqrt 5 }}\left( {x-\sqrt {{x^2}-5} } \right) \cdot \;{\log _5}\left( {x + \sqrt {{x^2}-5} } \right) + 1,5 = 0\)
ОТВЕТ: \(\dfrac{{13\sqrt 5 }}{5}.\)
\({\log _{\sqrt 5 }}\left( {x-\sqrt {{x^2}-5} } \right) \cdot {\log _5}\left( {x + \sqrt {{x^2}-5} } \right) + 1,5 = 0.\) Заметим, что: \(x-\sqrt {{x^2}-5} = \dfrac{{\left( {x-\sqrt {{x^2}-5} } \right)\left( {x + \sqrt {{x^2}-5} } \right)}}{{x + \sqrt {{x^2}-5} }} = \dfrac{{{x^2}-{x^2} + 5}}{{x + \sqrt {{x^2}-5} }} = \dfrac{5}{{x + \sqrt {{x^2}-5} }}.\) Тогда уравнение примет вид: \(2{\log _5}\dfrac{5}{{x + \sqrt {{x^2}-5} }} \cdot {\log _5}\left( {x + \sqrt {{x^2}-5} } \right) + 1,5 = 0\,\,\,\,\,\,\, \Leftrightarrow \) \( \Leftrightarrow \,\,\,\,\,\,\,2 \cdot \left( {{{\log }_5}5-{{\log }_5}\left( {x + \sqrt {{x^2}-5} } \right)} \right) \cdot {\log _5}\left( {x + \sqrt {{x^2}-5} } \right) + 1,5 = 0.\) Пусть \({\log _5}\left( {x + \sqrt {{x^2}-5} } \right) = t\). Тогда: \(2\left( {1-t} \right) \cdot t + 1,5 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,4{t^2}-4t-3 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{t = -\dfrac{1}{2},}\\{t = \dfrac{3}{2}.\,\,\,}\end{array}} \right.\) Возвращаясь к прежней переменной, получим: \(\left[ {\begin{array}{*{20}{c}}{{{\log }_5}\left( {x + \sqrt {{x^2}-5} } \right) = -\dfrac{1}{2},}\\{{{\log }_5}\left( {x + \sqrt {{x^2}-5} } \right) = \dfrac{3}{2}\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x + \sqrt {{x^2}-5} = \dfrac{1}{{\sqrt 5 }},\,}\\{x + \sqrt {{x^2}-5} = 5\sqrt {5.} }\end{array}} \right.} \right.\) Рассмотрим первое уравнение: \(x + \sqrt {{x^2}-5} = \dfrac{1}{{\sqrt 5 }}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\sqrt {{x^2}-5} = \dfrac{1}{{\sqrt 5 }}-x.\) Получили уравнение вида: \(\sqrt {f\left( x \right)} = g\left( x \right)\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{g\left( x \right) \ge 0,\,\,\,\,\,\,\,\,\,\,\,}\\{f\left( x \right) = {g^2}\left( x \right).}\end{array}} \right.\) \(\sqrt {{x^2}-5} = \dfrac{1}{{\sqrt 5 }}-x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{x^2}-5 = \dfrac{1}{5}-\dfrac{2}{{\sqrt 5 }}x + {x^2},}\\{\dfrac{1}{{\sqrt 5 }}-x \ge 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x = \dfrac{{13\sqrt 5 }}{5},}\\{x \le \dfrac{{\sqrt 5 }}{5}\,\,\,\,\,}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\emptyset .} \right.} \right.\) Рассмотрим второе уравнение: \(\sqrt {{x^2}-5} = 5\sqrt 5 -x\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{x^2}-5 = 125-10\sqrt 5 x + {x^2}}\\{5\sqrt 5 -x \ge 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\, \Leftrightarrow } \right.\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x = \dfrac{{13\sqrt 5 }}{5},}\\{x \le 5\sqrt 5 \,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,x = \dfrac{{13\sqrt 5 }}{5}.\) Ответ: \(\dfrac{{13\sqrt 5 }}{5}\).