Do \(\left|\left(x-y\right)^2+2\left(x^2+xy-4y\right)\right|\ge0;\forall x;y\) theo tính chất trị tuyệt đối
Mà \(\left|\left(x-y\right)^2+2\left(x^2+xy-4y\right)\right|=x^2+xy-4y\) (1)
\(\Rightarrow x^2+xy-4y\ge0\)
Ta có: \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0;\forall x;y\\x^2+xy-4y\ge0\left(cmt\right)\end{matrix}\right.\) (2)
\(\Rightarrow\left(x-y\right)^2+2\left(x^2+xy-4y\right)\ge0\)
\(\Rightarrow\left|\left(x-y\right)^2+2\left(x^2+xy-4y\right)\right|=\left(x-y\right)^2+2\left(x^2+xy-4y\right)\)
Nên (1) trở thành:
\(\left(x-y\right)^2+2\left(x^2+xy-4y\right)=x^2+xy-4y\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x^2+xy-4y\right)=0\) (3)
Từ (2);(3) \(\Rightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=0\\x^2+xy-4y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x^2+xy-4y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\x^2+x^2-4x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\2x\left(x-2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=y=0\\x=y=2\end{matrix}\right.\)
