Với mọi giá trị của x;y ta có:
\(\left(x-11+y\right)^2+\left(x-4-y\right)^2\ge0\)
Để \(\left(x-11+y\right)^2+\left(x-4-y\right)^2=0\) thì:
\(\left\{{}\begin{matrix}\left(x-11+y\right)^2=0\\\left(x-4-y\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y=11\\x-y=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+x-y=11+4\\y=x-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=15\\y=x-4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{15}{2}\\y=\dfrac{15}{2}-4=\dfrac{7}{2}\end{matrix}\right.\)
Chúc bạn học tốt!!!