\(A=2x^2+5y^2-2xy+2x+2y\)

\(=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(4y^2+2.2y.\frac{1}{2}+\frac{1}{4}\right)-1-\frac{1}{4}\)

\(=\left(x-y\right)^2+\left(x+1\right)^2+\left(2y+\frac{1}{2}\right)^2-\frac{5}{4}\)

Ta thấy: \(\left(x-y\right)^2\ge0;\left(x+1\right)^2\ge0;\left(2y+\frac{1}{2}\right)^2\ge0\forall x;y\)

\(\Rightarrow\left(x-y\right)^2+\left(x+1\right)^2+\left(2y+\frac{1}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)

\(\Rightarrow Min_A=-\frac{5}{4}\). 

\(M=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+3y^2-2\)

\(M=\left(x-y\right)^2+\left(x+1\right)^2+\left(y+1\right)^2+3y^2-2\ge-2\)

\(A=2x^2+5y^2-2xy+2y+2x\)

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

\(=\left(x+y+1\right)^2+\left(x-2y\right)^2-1\)

Ta thấy :(x + y +1)² ≥ 0 ∀ x,y

(x - 2y)² ≥ 0 ∀ x,y

⇒ (x + y +1)² +(x - 2y)² ≥ 0 ∀ x,y

⇔(x + y +1)² +(x - 2y)² -1 ≥ -1 ∀ x,y

⇔ A ≥ -1 ∀ x,y

Vậy GTNN của A là -1 \(\Leftrightarrow\left\{{}\begin{matrix}x+y+1=0\\x-2y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=-1\\x=2y\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-2}{3}\\y=\dfrac{-1}{3}\end{matrix}\right.\)

Đặt \(A=-2x^2-y^2-2xy+4x+2y+2\)

\(-A=2x^2+y^2+2xy-3x-2y-2\)

\(-A=\left(x^2+2xy+y^2\right)+x^2-4x-2y-2\)

\(-A=\left[\left(x+y\right)^2-2\left(x+y\right)+1\right]+\left(x^2-2x+1\right)-4\)

\(-A=\left(x+y-1\right)^2+\left(x-1\right)^2-4\)

Mà  \(\left(x+y-1\right)^2\ge0\forall x;y\)

       \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow-A\ge-4\)

\(\Leftrightarrow A\le4\)

Dấu "=" xảy ra khi :

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

Vậy  \(A_{Max}=4\Leftrightarrow\left(x;y\right)=\left(1;0\right)\)

Đặt  \(B=x^2-4xy+5y^2+10x-22y+27\)

\(B=\left(x^2-4xy+4y^2\right)+y^2+10x-22y+27\)

\(B=\left[\left(x-2y\right)^2+2\left(x-2y\right)\times5+25\right]+\)\(\left(y^2-2y+1\right)+1\)

\(B=\left(x-2y+5\right)^2+\left(y-1\right)^2+1\)

Mà  \(\left(x-2y+5\right)^2\ge0\forall x;y\)

       \(\left(y-1\right)^2\ge0\forall y\)

\(\Rightarrow B\ge1\)

Dấu "=" xảy ra khi :

\(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vậy  \(B_{Min}=1\Leftrightarrow\left(x;y\right)=\left(-3;1\right)\)

\(2xy+2x-5z=0\Rightarrow5z=2xy+2x\Rightarrow z=\frac{2}{5}xy+\frac{2}{5}x\)

\(A=x^2+2y^2+2xy+\frac{8}{5}y+z+2\)

\(A=x^2+2y^2+2xy+\frac{8}{5}y+\frac{2}{5}xy+\frac{2}{5}x+2\)

\(A=x^2+2y^2+\frac{12}{5}xy+\frac{2}{5}x+\frac{8}{5}y+2\)

\(A=x^2+\left(\frac{6y}{5}\right)^2+\left(\frac{1}{5}\right)^2+2.\frac{6}{5}xy+\frac{2}{5}x+\frac{12y}{25}+\frac{14}{25}y^2+\frac{28y}{25}+\frac{14}{25}+\frac{7}{5}\)

\(A=\left(x+\frac{6y}{5}+\frac{1}{5}\right)^2+\frac{14}{25}\left(y+1\right)^2+\frac{7}{5}\ge\frac{7}{5}\)

\(\Rightarrow A_{min}=\frac{7}{5}\) khi \(\left\{{}\begin{matrix}x=1\\y=-1\\z=0\end{matrix}\right.\)