\(x^2+y^2+1=xy+x+y\)
\(\Leftrightarrow x^2+y^2+1-xy-x-y=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(x^2-2xy+y^2\right)+\dfrac{1}{2}\left(x^2-2x+1\right)+\dfrac{1}{2}\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(x-y\right)^2+\dfrac{1}{2}\left(x-1\right)^2+\dfrac{1}{2}\left(y-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-1=0\\y-1=0\end{matrix}\right.\Leftrightarrow x=y=1\)
\(A=\left(x+y-3\right)^{2020}=\left(1+1-3\right)^{2020}=\left(-1\right)^{2020}=1\)
Đúng 1
Bình luận (0)
