$\text{Tìm min:}$ $H=2x^2 + y^2 -2xy +2y +2021$ $\text{ANH PUVI giúp iem với ạ}$ 09/11/2021 Bởi Clara $\text{Tìm min:}$ $H=2x^2 + y^2 -2xy +2y +2021$ $\text{ANH PUVI giúp iem với ạ}$
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 $MinH = 2019 \Leftrightarrow \left( {x;y} \right) = \left( { – 1; – 2} \right)$ Bình luận
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 $\min H = 2019 \Leftrightarrow (x;y)=(-1;-2)$ Bình luận
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 $MinH = 2019 \Leftrightarrow \left( {x;y} \right) = \left( { – 1; – 2} \right)$
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 $\min H = 2019 \Leftrightarrow (x;y)=(-1;-2)$