Question

[Đức Anh]

tìm giá trị nhỏ nhất: A=x ^{ 2  }  +5y ^{ 2  }  -4xy-2y+2x+2010 

Answer 1

[Lấp La Lấp Lánh]

\(A=x^2+5y^2-4xy-2y+2x+2010\)

\(=\left[x^2-2x\left(2y-1\right)+\left(2y-1\right)^2\right]+\left(y^2+2y+1\right)+2008\)

\(=\left(x-2y+1\right)^2+\left(y+1\right)^2+2008\ge2008\)

\(minA=2008\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\)

Answer 2

[Nguyễn Hoàng Minh]

\(A=\left[\left(x^2-4xy+4y^2\right)+2\left(x-2y\right)+1\right]+\left(y^2+2y+1\right)+2008\\ A=\left[\left(x-2y\right)^2+2\left(x-2y\right)+1\right]+\left(y+1\right)^2+2008\\ A=\left(x-2y+1\right)^2+\left(y+1\right)^2+2008\ge2008\\ A_{min}=2008\Leftrightarrow\left\{{}\begin{matrix}x=2y-1\\y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\)