a) Ta có A = -4x2 - 12x = -4x2 - 12x - 9 + 9 = -(2x + 3)2 + 9 \(\le9\)
Dấu "=" xảy ra <=> 2x + 3 = 0
<=> x = -1,5
Vậy Max A = 9 <=> x = -1,5
b) Ta có B = 7 - x2 - y2 - 2(x + y)
= -x2 - 2x - 1 - y2 - 2y - 1 + 9
= -(x + 1)2 - (y + 1)2 + 9 \(\le9\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+1=0\\y+1=0\end{cases}}\Leftrightarrow x=y=-1\)
Vậy Max B = 9 <=> x = y = -1
\(A=-\left(4x^2+12x\right)\)
\(A=-\left(4x^2+12x+9\right)+9\)
\(A=-\left(2x+3\right)^2+9\le9\)
\(< =>MAX:A=9\)dấu "=" xảy ra khi \(2x+3=0< =>x=-\frac{3}{2}\)
\(B=7-x^2-y^2-2x-2y\)
\(B=7-\left(x^2+2x\right)-\left(y^2+2y\right)\)
\(B=9-\left(x^2+2x+1\right)-\left(y^2+2y+1\right)\)
\(B=9-\left(x+1\right)^2-\left(y+1\right)^2\le9\)
\(< =>MAX:B=9\)dấu "=" xảy ra khi \(\hept{\begin{cases}x+1=0\\y+1=0\end{cases}\hept{\begin{cases}x=-1\\y=-1\end{cases}}}\)
A = -4x2 - 12x = -4(x2 + 3x + 9/4) + 9 = -4(x + 3/2)2 + 9 \(\le\)9
Dấu "=" xảy ra <=> x + 3/2 = 0 <=> x= -3/2
Vậy MaxA = 9 <=> x = -3/2
B = 7 - x2 - y2 - 2(x + y)
B =9 - (x2 + 2x + 1) - (y2 + 2y + 1) = 9 - (x + 1)2 - (y + 1)2 \(\le\)9
Dấu "=" xảy ra <==> x + 1 = 0 và y + 1 = 0 <=> x = y = -1
Vậy MaxB = 9 <=> x = y = -1