\(x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=4\)
\(\Rightarrow\left(x^2-2\cdot x\cdot\dfrac{1}{x}+\dfrac{1}{x^2}\right)+\left(y^2-2\cdot y\cdot\dfrac{1}{y}+\dfrac{1}{y^2}\right)+2+2=4\)
\(\Rightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\)
Vì \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{x}\right)^2\ge0\\\left(y-\dfrac{1}{y}\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{x}\right)^2=0\\\left(y-\dfrac{1}{y}\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\y-\dfrac{1}{y}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{x}\\y=\dfrac{1}{y}\end{matrix}\right.\)
\(\Rightarrow x^2=y^2=1\)
\(\Rightarrow\left[{}\begin{matrix}x=y=1\\x=y=-1\\x=1;y=-1\\x=-1;y=1\end{matrix}\right.\)
Vậy...............................................................
\(x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=4\)
\(\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)=0\)
\(\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\)
lm như bt
