H=\(x^6-2x^3+x^2-2x+2\)

\(=x^6+2x^5+3x^4+2x^2-2x^5-4x^4-6x^3-4x^2-4x+x^4+2x^3+3x^2+2x+2\)

\(=x^2\left(x^4+2x^3+3x^2+2\right)-2x\left(x^4+2x^3+3x^2+2\right)+\left(x^4+2x^3+3x^2+2\right)\)

\(=\left(x^2-2x+1\right)\left(x^4+2x^3+3x^2+2\right)\)

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

\(=\left(x-1\right)^2\left(x^2+1\right)\left[\left(x+1\right)^2+1\right]\text{≥}0\)

Vì \(\left\{{}\begin{matrix}\left(x-1\right)^2\text{≥}0\\\left(x^2+1\right)\text{≥}1\\\left(x+1\right)^2+1\text{≥}1\end{matrix}\right.\)

⇒ MinH=0 ⇔ \(x=1\)

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