\(a,A=3-4x-x^2\)
\(=-\left(x^2+4x+4\right)+7\)
\(=-\left(x+2\right)^2+7\)
Với mọi giá trị của x ta có:
\(\left(x+2\right)^2\ge0\Rightarrow-\left(x+2\right)^2\le0\)
\(\Rightarrow-\left(x+2\right)^2+7\le7\)
Vậy Max A = 7 khi \(x+2=0\Rightarrow x=-2\)
\(b,B=2x-x-3x^2=x-3x^2\)
\(=-3\left(x^2-\dfrac{1}{3}x+\dfrac{1}{36}\right)+\dfrac{1}{12}\)
\(=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{1}{12}\)
Với mọi giá trị của x ta có:
\(\left(x-\dfrac{1}{6}\right)^2\ge0\Rightarrow-3\left(x-\dfrac{1}{6}\right)^2\le0\)
\(\Rightarrow-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{1}{12}\le\dfrac{1}{12}\)
Vậy Max B = \(\dfrac{1}{12}\) khi \(x-\dfrac{1}{6}=0\Rightarrow x=\dfrac{1}{6}\)
\(c,C=2-x^2-y^2-2\left(x+y\right)=2-x^2-y^2-2x-2y\)\(=4-\left(x^2+2x+1\right)-\left(y^2+2y+1\right)\)
\(=4-\left(x+1\right)^2-\left(y+1\right)^2\)
Với mọi giá trị của x , ta có:
\(\left(x+1\right)^2\ge0;\left(y+1\right)^2\ge0\)
\(\Rightarrow4-\left(x+1\right)^2-\left(y+1\right)^2\le4\)
Vậy Max C = 4 khi \(\left\{{}\begin{matrix}x+1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-1\end{matrix}\right.\)
\(d,D=-x^2+4x-9=-\left(x^2-4x+4\right)-5\) \(=-\left(x-2\right)^2-5\)
Với mọi giá trị của x ta có:
\(\left(x-2\right)^2\ge0\Rightarrow-\left(x-2\right)^2\le0\)
\(\Rightarrow-\left(x-2\right)^2-5\le-5\)
Vậy Max D = -5 khi \(x-2=0\Rightarrow x=2\)
\(e,E=-x^2+4x-y^2-12y+47\)
\(=-\left(x^2-4x+4\right)-\left(y^2+12y+36\right)+87\)
\(=-\left(x-2\right)^2-\left(y+6\right)^2+87\)
Với mọi giá trị của x ta có:
\(-\left(x-2\right)^2\le0;-\left(y+6\right)\le0\)
\(\Rightarrow-\left(x-2\right)^2-\left(y+6\right)^2+87\le87\)
Vậy Max E = 87
Để E = 87 thì \(\left\{{}\begin{matrix}x-2=0\\y+6=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=-6\end{matrix}\right.\)
\(f,F=-x^2-x-y^2-3y+13\)
\(=-\left(x^2+x+\dfrac{1}{4}\right)-\left(y^2+3y+\dfrac{9}{4}\right)+\dfrac{31}{2}\)
\(=-\left(x+\dfrac{1}{2}\right)^2-\left(y+\dfrac{3}{2}\right)^2+\dfrac{31}{2}\)
Với mọi giá trị của x ta có:
\(-\left(x+\dfrac{1}{2}\right)^2\le0;-\left(y+\dfrac{3}{2}\right)^2\le0\)
\(\Rightarrow-\left(x+\dfrac{1}{2}\right)^2-\left(y+\dfrac{3}{2}\right)^2+\dfrac{31}{2}\le\dfrac{31}{2}\)
Vậy Max F = \(\dfrac{31}{2}\) khi \(\left\{{}\begin{matrix}x+\dfrac{1}{2}=0\\y+\dfrac{3}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
