Xin phép được sửa đề :3
\(x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=4\)
\(\Leftrightarrow\left(x^2+\dfrac{1}{x^2}\right)+\left(y^2+\dfrac{1}{y^2}\right)=4\)
Áp dụng BĐT Cô - Si cho hai số không âm ta có :
\(\left\{{}\begin{matrix}x^2+\dfrac{1}{x^2}\ge2\sqrt{x^2\times\dfrac{1}{x^2}}=2\\y^2+\dfrac{1}{y^2}\ge2\sqrt{y^2\times\dfrac{1}{y^2}}=2\end{matrix}\right.\)
\(\Rightarrow x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}\ge4\)
Dấu \("="\) xảy ra khi :
\(\left\{{}\begin{matrix}x^2=\dfrac{1}{x^2}\Leftrightarrow x^4=1\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\\y^2=\dfrac{1}{y^2}\Leftrightarrow y^4=1\Leftrightarrow\left[{}\begin{matrix}y=-1\\y=1\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(-1;-1\right)\) hoặc \(\left(x;y\right)=\left(1;1\right)\)
Wish you study well !!
