\[\dfrac{1}{x} + \dfrac{1}{{\sqrt {1 - {x^2}} }} = 2\sqrt 2 \]
ĐK: \[x \in \left( { - 1;0} \right) \cup \left( {0;1} \right)\]
Đặt: \[x = \sin t,t \in \left( { - \dfrac{\pi }{2};0} \right) \cup \left( {0;\dfrac{\pi }{2}} \right)\]
\[\begin{array}{l}
pt \Leftrightarrow \dfrac{1}{{\sin t}} + \dfrac{1}{{\sqrt {1 - {{\sin }^2}t} }} = 2\sqrt 2 \\
\Leftrightarrow \sin t + \cos t = 2\sqrt 2 \sin t\cos t\\
\Leftrightarrow \sin \left( {t + \dfrac{\pi }{4}} \right) = \sin 2t\\
\Leftrightarrow \left[ \begin{array}{l}
t = \dfrac{\pi }{4} + k2\pi \\
t = - \dfrac{\pi }{{12}} + k\dfrac{{2\pi }}{3}
\end{array} \right.\\
t \in \left( { - \dfrac{\pi }{2};0} \right) \cup \left( {0;\dfrac{\pi }{2}} \right) \Rightarrow \left[ \begin{array}{l}
t = \dfrac{\pi }{4}\\
t = - \dfrac{\pi }{{12}}
\end{array} \right.\\
\bullet t = \dfrac{\pi }{4} \Rightarrow x = \dfrac{{\sqrt 2 }}{2}(TM)\\
\bullet t = - \dfrac{\pi }{{12}} \Rightarrow x = \dfrac{{ - \sqrt 6 + \sqrt 2 }}{4}(TM)
\end{array}\]
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