\(x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\)
\(\Leftrightarrow\)\(\left(x^2+\frac{1}{x^2}-2\right)+\left(y^2+\frac{1}{y^2}-2\right)=0\)
\(\Leftrightarrow\)\(\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x-\frac{1}{x}=0\\y-\frac{1}{y}=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x^2=1\\y^2=1\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=\pm1\\y=\pm1\end{cases}}\)
\(^{x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4}\)
\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)+\left(y^2+\frac{1}{y^2}\right)-2-2=0\)
\(\Leftrightarrow\left(x^2-2.x.\frac{1}{x}+\frac{1}{x^2}\right)+\left(y^2-2.y.\frac{1}{y}+\frac{1}{y^2}\right)=0\)
\(\Leftrightarrow\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)
Mặt khác: \(\left(x-\frac{1}{x}\right)^2\ge0\)\(\forall\)x\(\ne\)0
\(\left(y-\frac{1}{y}\right)^2\ge0\)\(\forall\)y \(\ne\)0
Từ hai điều trên \(\Rightarrow\)\(\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2\ge0\)\(\forall\)x,y \(\ne\)0
Dấu "=" xảy ra
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-\frac{1}{x}\right)^2=0\\\left(y-\frac{1}{y}\right)^2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{x}=0\\y-\frac{1}{y}=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x^2-1=0\\y^2-1=0\end{cases}}\)(vì x,y khác 0)
\(\Leftrightarrow\hept{\begin{cases}x^2=1\\y^2=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\pm1\\y=\pm1\end{cases}}\)
Vậy.....
Ta có: \(x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\)
\(\Leftrightarrow\left(x^2-2+\frac{1}{x^2}\right)+\left(y^2-2+\frac{1}{y^2}\right)=0\)
\(\Leftrightarrow\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{x}=0\\y-\frac{1}{y}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\pm1\\y=\pm1\end{cases}}\)
