\(A=x^2+y^2+xy-3x-3y+2-18\)

\(=\left(x^2+\dfrac{y^2}{4}+\dfrac{9}{4}+xy-3x-\dfrac{3y}{2}\right)+\dfrac{3}{4}\left(y^2-2y+1\right)+2015\)\(=\left(x+\dfrac{y}{2}-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\left(y-1\right)^2+2015\ge2015\)

\(A_{min}=2015\) khi \(\left(x;y\right)=\left(1;1\right)\)

Đặt  \(A=x^2+y^2+xy+3x+3y+2018\)

\(4.A=4x^2+4y^2+4xy+12x+12y+8072\)

\(4.A=\left(4x^2+4xy+y^2\right)+3y^2+12x+12y+8072\)

\(4.A=\left[\left(2x+y\right)^2+2\left(2x+y\right).3+9\right]+3\left(y^2+2y+1\right)+8060\)

\(4.A=\left(2x+y+3\right)^2+3\left(y+1\right)^2+8060\)

Mà  \(\left(2x+y+3\right)^2\ge0\forall x;y\)

       \(\left(y+1\right)^2\ge0\forall y\)\(\Rightarrow3\left(y+1\right)^2\ge0\forall y\)

\(\Rightarrow4.A\ge8060\)

\(\Leftrightarrow A\ge2015\)

Dấu "=" xảy ra khi : 

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

Vậy ...

Đặt \(A=x^2-xy+y^2-3x-3y+2029\)

\(\Leftrightarrow2A=2x^2-2xy+2y^2-6x-6y+4058\)

\(\Leftrightarrow2A=\left(x^2-2xy+y^2\right)+\left(x^2-6x+9\right)+\left(y^2-6y+9\right)+4040\)

\(\Leftrightarrow2A=\left(x-y\right)^2+\left(x-3\right)^2+\left(y-3\right)^2+4040\ge4040\forall x;y\)

\(\Leftrightarrow A\ge\dfrac{4040}{2}=2020\forall x;y\)

Dấu " = " xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\x-3=0\\y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=3\\y=3\end{matrix}\right.\) \(\Leftrightarrow x=y=3\)

Vậy GTNN của b/t trên là : \(2020\Leftrightarrow x=y=3\)

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Tìm GTNN của - K2PI – TOÁN THPT | Chia sẻ Tài liệu, đề thi, hỗ trợ giải toán

\(A=x^2+y^2+xy+3x+3y+2018\)

\(\Leftrightarrow2A=2x^2+2y^2+2xy+6x+6y+4036\)

\(=\left(x^2+2xy+y^2\right)+\left(x^2+6x+9\right)+\left(y^2+6y+9\right)+4018\)

\(=\left(x+y\right)^2+\left(x+3\right)^2+\left(y+3\right)^2+4018\)

\(\Rightarrow A=\dfrac{\left(x+y\right)^2+\left(x+3\right)^2+\left(y+3\right)^2}{2}+2009\)

Ta có : \(\left\{{}\begin{matrix}\left(x+y\right)^2\ge0\\\left(x+3\right)^2\ge0\\\left(y+3\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\dfrac{\left(x+y\right)^2+\left(x+3\right)^2+\left(y+3\right)^2}{2}+2009\ge2009\)

Dấu = xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=0\\\left(x+3\right)^2=0\\\left(y+3\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow x=y=-3\)

Vậy \(Min_A=2009\Leftrightarrow x=y=-3\)