1, \(C=2x\left(6-2x\right)=12x-4x^2\)
\(=-4\left(x^2-\dfrac{3}{2}x.2+\dfrac{9}{4}-\dfrac{9}{4}\right)\)
\(=-4\left(x-\dfrac{3}{2}\right)^2+9\le9\)
Dấu " = " khi \(-4\left(x-\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{3}{2}\)
Vậy \(MAX_C=-9\) khi \(x=\dfrac{3}{2}\)
2, \(D=-9x^2-6x-2=-\left(9x^2+6x+1-3\right)\)
\(=-\left(3x+1\right)^2+3\le3\)
Dấu " = " khi \(-\left(3x+1\right)^2=0\Leftrightarrow x=\dfrac{-1}{3}\)
Vậy \(MAX_D=3\) khi \(x=\dfrac{-1}{3}\)
\(C=2x\left(6-2x\right)=12x-4x^2=-\left(4x^2-12x+9\right)-9\)
\(\Leftrightarrow-\left(2x-3\right)^2-9\)
Ta có : \(\left(2x+3\right)^2\ge0\)
\(\Leftrightarrow-\left(2x-3\right)^2\le0\)
\(\Leftrightarrow-\left(2x-3\right)^2-9\le-9\)
Vậy \(Max_C=-9\Leftrightarrow x=\dfrac{3}{2}.\)
\(D=-9x^2-6x-2\)
\(=-\left(9x^2+6x+2\right)=-\left(3x+1\right)+1\le1\)
Vậy \(Max_D=1\Leftrightarrow x=-\dfrac{1}{3}.\)
Z 
