1: \(A=-4x^2+12x+3\)
\(=-4\left(x^2-3x-\dfrac{3}{4}\right)\)
\(=-4\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-3\right)\)
\(=-4\left(x-\dfrac{3}{2}\right)^2+12\le12\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)
2: Ta có: \(M=-x^2+6x+3\)
\(=-\left(x^2-6x+9-12\right)\)
\(=-\left(x-3\right)^2+12\le12\forall x\)
Dấu '=' xảy ra khi x=3
1) \(A=-\left(4x^2-12x+9\right)+12=-\left(2x-3\right)^2+12\le12\)
\(maxA=12\Leftrightarrow x=\dfrac{3}{2}\)
2) \(M=6x-x^2+3=-\left(x^2-6x+9\right)+12=-\left(x-3\right)^2+12\le12\)
\(maxM=12\Leftrightarrow x=3\)
3) \(E=-\left(x^2-2x+1\right)-\left(y^2+6y+9\right)+8=-\left(x-1\right)^2-\left(y+3\right)^2+8\le8\)
\(maxE=8\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
4) \(B=-\left(4x^2-12x+9\right)-\left(y^2+8y+16\right)+26=-\left(2x-3\right)^2-\left(y+4\right)^2+26\le26\)
\(maxB=26\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-4\end{matrix}\right.\)
