Đáp án:

$\min H = 2019 \Leftrightarrow (x;y)=(-1;-2)$

Giải thích các bước giải:

$\quad H= 2x^2 + y^2 - 2xy + 2y + 2021$

$\to H = \dfrac12(4x^2 - 4xy + y^2) +\dfrac12(y^2 + 4y + 4) + 2019$

$\to H =\dfrac12(2x -y)^2 +\dfrac12(y +2)^2 + 2019$

Ta có:

$\quad \begin{cases}(2x-y)^2 \geq 0\quad \forall x;y\\(y+2)^2\geq 0\quad \forall y\end{cases}$

Do đó:

$\quad \dfrac12(2x -y)^2 +\dfrac12(y +2)^2 + 2019\geq 2019$

Hay $H \geq 2019$

Dấu $=$ xảy ra $\Leftrightarrow \begin{cases}2x - y = 0\\y +2 = 0\end{cases}\Leftrightarrow \begin{cases}x = -1\\y = -2\end{cases}$

Vậy $\min H = 2019 \Leftrightarrow (x;y)=(-1;-2)$

Giải thích các bước giải:

 Ta có:

$\begin{array}{l}
H = 2{x^2} + {y^2} - 2xy + 2y + 2021\\
 = \dfrac{1}{2}\left( {4{x^2} - 4xy + {y^2}} \right) + \dfrac{1}{2}{y^2} + 2y + 2021\\
 = \dfrac{1}{2}{\left( {2x - y} \right)^2} + \dfrac{1}{2}\left( {{y^2} + 4y + 4} \right) + 2019\\
 = \dfrac{1}{2}{\left( {2x - y} \right)^2} + \dfrac{1}{2}{\left( {y + 2} \right)^2} + 2019
\end{array}$

Mà:

$\begin{array}{l}
\left\{ \begin{array}{l}
{\left( {2x - y} \right)^2} \ge 0\\
{\left( {y + 2} \right)^2} \ge 0
\end{array} \right.,\forall x,y\\
 \Rightarrow \dfrac{1}{2}{\left( {2x - y} \right)^2} + \dfrac{1}{2}{\left( {y + 2} \right)^2} \ge 0\\
 \Rightarrow \dfrac{1}{2}{\left( {2x - y} \right)^2} + \dfrac{1}{2}{\left( {y + 2} \right)^2} + 2019 \ge 2019\\
 \Rightarrow H \ge 2019
\end{array}$

Dấu bằng xảy ra

$\begin{array}{l}
 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {2x - y} \right)^2} = 0\\
{\left( {y + 2} \right)^2} = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
2x - y = 0\\
y + 2 = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
y = 2x\\
y =  - 2
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x =  - 1\\
y =  - 2
\end{array} \right.
\end{array}$

Vậy $MinH = 2019 \Leftrightarrow \left( {x;y} \right) = \left( { - 1; - 2} \right)$