a, Ta có: \(-\left(x-3\right)^2\le0,\forall x\)
\(\Leftrightarrow-\left(x-3\right)^2-7\le-7\)
Vậy max A là \(-7\Leftrightarrow x=3\)
b, Ta có \(B=-x^2-2x-5=-\left(x+1\right)^2-4\le4\)
Dấu \("="\Leftrightarrow x=-1\)
c, Ta có \(C=-4x^2-4x+9=-\left(2x+1\right)^2+10\le10\)
Dấu \("="\Leftrightarrow x=-\dfrac{1}{2}\)
d, Ta có:\(D=-3y^2-6y+1\\ =-3\left(y^2+2y-\dfrac{1}{3}\right)=-3\left[\left(y+1\right)^2-\dfrac{4}{3}\right]=-3\left(y+1\right)^2+4\le4\)
Dấu \("="\Leftrightarrow y=-1\)
a:Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Leftrightarrow-\left(x-3\right)^2\le0\forall x\)
\(\Leftrightarrow-\left(x-3\right)^2-7\le-7\forall x\)
Dấu '=' xảy ra khi x=3
c: Ta có: \(C=-4x^2-4x+9\)
\(=-\left(4x^2+4x-9\right)\)
\(=-\left(4x^2+4x+1-10\right)\)
\(=-\left(2x+1\right)^2+10\le10\forall x\)
Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)