anh ko biết nha em yêu của anh

\(A=2x^2+9y^2-6xy-6x-12y+2046\)

\(=\left[\left(x^2-6xy+9y^2\right)+\left(4x-12y\right)+4\right]-4+\left(x^2-10x+25\right)-25+2046\)

\(=\left[\left(x-3y\right)^2+4\left(x-3y\right)+4\right]+\left(x-5\right)^2-4-25+2046\)

\(=\left(x-3y+2\right)^2+\left(x-5\right)^2+2017\ge2017\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-3y+2=0\\x-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=\frac{7}{3}\\x=5\end{cases}}}\)

Vậy \(A_{min}=2017\) tại \(x=5;y=\frac{7}{3}\)

\(C=2x^2+9y^2-6xy-2x+2018\)

\(=\left(x^2-6xy+9y^2\right)+\left(x^2-2x+1\right)+2017\)

\(=\left(x-3y\right)^2+\left(x-1\right)^2+2017\)

Nhận xét :

\(\left\{{}\begin{matrix}\left(x-3y\right)^2\ge0\\\left(x-1\right)^2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left(x-3y\right)^2+\left(x-1\right)^2\ge0\)

\(\Leftrightarrow\left(x-3y\right)^2+\left(x-1\right)^2+2017\ge2017\)

\(\Leftrightarrow C\ge2017\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3y\right)^2=0\\\left(x-1\right)^2=0\end{matrix}\right.\)

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

Vậy \(C_{Min}=2017\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{3}\end{matrix}\right.\)

\(D=x^2-2xy+6y^2-12x+2y+45\)

\(=\left(x^2-2xy+y^2\right)-\left(12x+12y\right)-10y+5y^2+45\)

\(=\left(x-y\right)^2-12\left(x-y\right)+36+\left(5y^2-10y+5\right)+4\)

\(=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)

Nhận xét :

\(\left\{{}\begin{matrix}\left(x-y-6\right)^2\ge0\\5\left(y-1\right)^2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left(x-y-6\right)+5\left(y-1\right)^2+4\ge4\)

\(\Leftrightarrow D\ge4\)

Dấu "=" xảy ra khi \(\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

Vậy \(D_{Min}=4\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)