Ta có:
\(x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=4\)
\(\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)+4=4\)
\(\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\)
Vì \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{x}\right)^2\ge0\forall x\\\left(y-\dfrac{1}{y}\right)\ge0\forall y\end{matrix}\right.\)
Dấu "="⇔ \(\left\{{}\begin{matrix}x=\dfrac{1}{x}\\y=\dfrac{1}{y}\end{matrix}\right.\)
\(\Leftrightarrow x^2=y^2=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y=1\\x=y=-1\\x=1,y=-1\\x=-1,y=1\end{matrix}\right.\)
Thay vào phương trình 1
⇒ \(x=y=1\)
Đúng 3
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