Question

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Tính x.y

 

Answer 1

\(2x^2+\dfrac{1}{x^2}+\dfrac{y^2}{4}=4\\ \Leftrightarrow\left(x^2+xy+\dfrac{y^2}{4}\right)+\left(x^2-2+\dfrac{1}{x^2}\right)=xy+2\\ \Leftrightarrow\left(x+\dfrac{y}{2}\right)^2+\left(x-\dfrac{1}{x}\right)^2=xy+2\)

Vì \(\left(x+\dfrac{y}{2}\right)^2+\left(x-\dfrac{1}{x}\right)^2\ge0\)

\(\Leftrightarrow xy+2\ge0\Leftrightarrow xy\ge-2\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{y}{2}\\x=\dfrac{1}{x}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm2\end{matrix}\right.\)

Vậy pt có nghiệm \(\left(x;y\right)=\left(\pm1;\pm2\right)\)