Đáp án:
Giải thích các bước giải:
$\text{Ta có D=-x²-y²+xy+2x+2y}$
$\text{⇒-2D=2x²+2y²-2xy-2x-2y}$
$\text{⇒-2D=(x²-2x+1)+(x²-2xy+y²)+(y²-2y+1)-2-2}$
$\text{⇒-2D=(x²-2x1+1²)+(x²-2xy+y²)+(y²-2y1+1²)-2}$
$\text{⇒-2D=(x-1)²+(x-y)²+(y-1)²-2}$
$\text{⇒D=$\frac{(x-1)²+(x-y)²+(y-1)²-2}{-2}$ }$
$\text{⇒D=$\frac{(x-1)²}{-2}$+$\frac{(x-y)²}{-2}$+$\frac{+(y-1)²}{-2}$+1}$
$\text{⇒D=-$\frac{(x-1)²}{2}$-$\frac{(x-y)²}{2}$-$\frac{(y-1)²}{2}$+1}$
$\text{Vì $\begin{cases} -\frac{(x-1)²}{2}≤0\\-\frac{(x-y)²}{2}≤0\\-\frac{(y-1)²}{2}≤0 \end{cases}$}$
$\text{⇒D=-$\frac{(x-1)²}{2}$-$\frac{(x-y)²}{2}$-$\frac{(y-1)²}{2}$+1≤1 }$
$\text{Dấu "=" xảy ra ⇔$\begin{cases} x-1=0\\x-y=0\\y-1=0 \end{cases}$⇔x=y=1}$
$\text{Vậy Max D = 1 ⇔x=y=1}$
