Question

tìm cặp số x,y thỏa mãn

\(x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=4\)

Answer 1

ĐKXĐ: \(x;y\ne0\)

\(\Leftrightarrow\left(x^2+\dfrac{1}{x^2}-2\right)+\left(y^2+\dfrac{1}{y^2}-2\right)=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\y-\dfrac{1}{y}=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=1\\y^2=1\end{matrix}\right.\)

\(\Rightarrow\left(x;y\right)=\left(\pm1;\pm1\right)\)