\(A=-\left(x^2-xy+\dfrac{y^2}{4}\right)+2\left(x-\dfrac{y}{2}\right)-\dfrac{3y^2}{4}+3y\)
\(=-\left(x-\dfrac{y}{2}\right)^2+2\left(x-\dfrac{y}{2}\right)-1-\dfrac{3}{4}\left(y^2-4y+4\right)+4\)
\(=-\left(x-\dfrac{y}{2}-1\right)^2-\dfrac{3}{4}\left(y-2\right)^2+4\)
Do \(\left\{{}\begin{matrix}-\left(x-\dfrac{y}{2}-1\right)^2\le0\\-\dfrac{3}{4}\left(y-2\right)^2\le0\end{matrix}\right.\) ; \(\forall x;y\)
\(\Rightarrow A\le4\)
\(A_{max}=4\) khi \(\left\{{}\begin{matrix}x-\dfrac{y}{2}-1=0\\y-2=0\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(2;2\right)\)
\[ A = -x^2 - y^2 + xy + 2x + 2y \]
\[ A = -x^2 - y^2 + xy + 2x + 2y. \]
\[ A = -(x^2 - 2x) - (y^2 - 2y) + xy. \]
- \(x^2 - 2x = (x - 1)^2 - 1\)
- \(y^2 - 2y = (y - 1)^2 - 1\)
\[
A = -((x - 1)^2 - 1) - ((y - 1)^2 - 1) + xy
\]\[
= -(x - 1)^2 - (y - 1)^2 + 2 + xy.
\]\[
A = - (x - 1)^2 - (y - 1)^2 + xy + 2.
\]
- \(x - 1 = 0 \Rightarrow x = 1\)
- \(y - 1 = 0 \Rightarrow y = 1\)\[
A = - (1 - 1)^2 - (1 - 1)^2 + (1)(1) + 2
\]\[
= 0 + 1 + 2 = 3.
\]
Giá trị lớn nhất của biểu thức \(A = -x^2 - y^2 + xy + 2x + 2y\) là \(3\), đạt được khi \(x = 1\) và \(y = 1\).
