Ta có:
\(x^2+2y^2+2xy-4y+4=0\)
\(\Rightarrow\left(x^2+2xy+y^2\right)+\left(y^2-4y+4\right)=0\)
\(\Rightarrow\left(x+y\right)^2+\left(y-2\right)^2=0\)
Vì \(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\\\left(y-2\right)^2\ge0\end{matrix}\right.\) với mọi x,y
\(\Rightarrow\left(x+y\right)^2+\left(y-2\right)^2\ge0\) với mọi x,y
Mà \(\left(x+y\right)^2+\left(y-2\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=0\\y-2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-y\\y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\)
Thay x = -2 và y = 2 vào B, ta có:
\(B=\dfrac{x^2-7xy+52}{x-y}\)
\(B=\dfrac{\left(-2\right)^2-7.\left(-2\right).2+52}{\left(-2\right)-2}\)
\(B=\dfrac{84}{-4}=-21\)
