Cho các số thực dương \(x,y \ne 1\) và thỏa mãn \({\log _x}y = {\log _y}x,{\log _x}(x - y) = {\log _y}(x + y)\).

* Hướng dẫn giải

ĐK: \(x > y > 0,x,y \ne 1\)

Ta có:

\(\begin{array}{l}
\left\{ \begin{array}{l}
{\log _x}y = {\log _y}x\\
{\log _x}(x - y) = {\log _y}(x + y)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{\log _x}y = \frac{1}{{{{\log }_x}y}}\\
{\log _x}(x - y) = {\log _y}(x + y)
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
{\log _x}y =  \pm 1\\
{\log _x}(x - y) = {\log _y}(x + y)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
y = x(ktm)\\
y = \frac{1}{x}
\end{array} \right.\\
{\log _x}(x - y) = {\log _y}(x + y)
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
y = \frac{1}{x}\\
{\log _x}(x - y) = {\log _{{x^{ - 1}}}}(x + y)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
y = \frac{1}{x}\\
{\log _x}(x - y) + {\log _x}(x + y) = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
y = \frac{1}{x}\\
{\log _x}\left( {{x^2} - {y^2}} \right) = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
xy = 1\\
{x^2} - {y^2} = 1
\end{array} \right. \Leftrightarrow {x^2} + xy - {y^2} = 1 + 1 = 2
\end{array}\)