Ta có:

\(3-S=\left(x^2+4y^2+9z^2\right)-\left(2x+4y+6z\right)\)

\(\Rightarrow3-S=\left(x^2-2x+1\right)+\left(4y^2-4y+1\right)+\left(9z^2-6z+1\right)-3\)

\(\Rightarrow6-S=\left(x-1\right)^2+\left(2y-1\right)^2+\left(3z-1\right)^2\ge0\)

\(\Rightarrow S\le6\)

\(S_{max}=6\) khi \(\left\{{}\begin{matrix}x-1=0\\2y-1=0\\3z-1=0\end{matrix}\right.\) \(\Leftrightarrow\left(x;y;z\right)=\left(1;\dfrac{1}{2};\dfrac{1}{3}\right)\)

\(B=2x+12y+6z-x^2-4y^2-z^2-18\)

\(B=-\left(x^2-2x+1\right)-\left[\left(2y\right)^2-12y+9\right]-\left(z^2-6z+9\right)\)

\(B=-\left(x-1\right)^2-\left(2y-3\right)^2-\left(z-3\right)^2\)

Vì \(-\left(x-1\right)^2< 0\)Với mọi x

\(-\left(2y-3\right)^2< 0\)Với mọi y

\(-\left(z-3\right)^2< 0\)Với mọi z

Nên \(-\left(x-1\right)^2-\left(2y-3\right)^2-\left(z-3\right)^2< 0\)Với mọi x, y, z

Vậy GTLN của B \(\Leftrightarrow-\left(x-1\right)^2-\left(2y-3\right)^2-\left(z-3\right)^2=0\)

\(\left\{{}\begin{matrix}-\left(x-1\right)^2=0\\-\left(2y-3\right)^2=0\\-\left(z-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1,5\\z=3\end{matrix}\right.\)

\(-\left(2x^2+y^2+2xy-4x-2y-5\right)\\ \\ =-\left(x^2+2x\left(y-1\right)+\left(y^2-2y+1\right)+\left(x^2-2x+1\right)-7\right)\\ =-\left(x^2+2x\left(y-1\right)+\left(y-1\right)^2+\left(x-1\right)^2-7\right)\\ =-\left(\left(x+y-1\right)^2+\left(x-1\right)^2-7\right)\\ =-\left(x+y-1\right)^2-\left(x-1\right)^2-7\)

\(\left(x+y-1\right)^2\ge0\\ \Rightarrow-\left(x+y-1\right)^2\le0\\ \left(x-1\right)^2\ge0\\ \Rightarrow-\left(x+y-1\right)^2-\left(x-1\right)^2\le0\\ \Rightarrow-\left(x+y-1\right)^2-\left(x-1\right)^2-7\le-7\)

Max A = -7 khi x=1 ; y=0

B) TT

Ta có:

\(x^2+y^2+z^2-2x+4y=6z-14\\ \Leftrightarrow\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+\left(z^2-6z+9\right)=0\\ \Leftrightarrow\left(x-1\right)^2+\left(y+2\right)^2+\left(z-3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\\z=3\end{matrix}\right.\)

Thay vào p ta có: \(p=1^{2021}+\left(-2\right)^2+3=1+4+3=8\)

\(x^2+y^2+z^2-2x+4y=6z-14\)

\(\leftrightarrow (x^2-2x+1)+(y^2+4y+4)+(z^2-6z+9)=0\)

\(\leftrightarrow (x-1)^2+(y+2)^2+(z-3)^2=0\)

Ta có \(\begin{cases} (x-1)^2\ge 0\\(y+2)^2\ge 0\\(z-3)^2\ge 0\end{cases}\)

\(\to (x-1)^2+(y+2)^2+(z-3)0^2\ge 0\)

\(\to\) Dấu "=" xảy ra khi \(\begin{cases}x-1=0\\y+2=0\\z-3=0\end{cases}\)

\(\leftrightarrow \begin{cases}x=1\\y=-2\\z=3\end{cases}\)

Thay \(x=1;y=-2;z=3\) vào P

\(P=1^{2021}+(-2)^2+3=1+4+3=8\)

Vậy \(P=8\)

\(x^2+y^2+z^2+2x-4y+6z=-14\)

\(\Leftrightarrow\left(x^2+2x+1\right)+\left(y^2-4y+4\right)+\left(z^2+6z+9\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(y-2\right)^2+\left(z+3\right)^2=0\)

\(\Leftrightarrow\begin{cases}x+1=0\\y-2=0\\z+3=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=-1\\y=2\\z=-3\end{cases}\)

\(\Rightarrow x+y+z=-1+2-3=-2\)