Question

Tìm x,y thỏa mãn:

a)\(x^2+y^2=4-\frac{1}{x^2}-\frac{1}{y^2}\)

b)x²+2y²-2xy-2x+4y+2=0

Answer 1

a/ \(xy\ne0\)

\(\Leftrightarrow x^2-2+\frac{1}{x^2}+y^2-2+\frac{1}{y^2}=0\)

\(\Leftrightarrow x^2-2.x.\frac{1}{x}+\frac{1}{x^2}+y^2-2.y.\frac{1}{y}+\frac{1}{y^2}=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{x}\\y=\frac{1}{y}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\pm1\\y=\pm1\end{matrix}\right.\)

b/

\(x^2+y^2+1-2xy-2x+2y+y^2+2y+1=0\)

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

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