Câu hỏi của Anh Nguyễn Quang

Question

Tìm min:

A=x2+2y2+3z2-2xy+2xz-2x-2y-8z+2010

Giúp mk với, mk cần gấp

Phan Công Bằng · Accepted Answer

$A=x^2+2y^2+3z^2-2xy+2xz-2x-2y-8z+2010$

$=x^2-2x\left(y-z+1\right)+\left(y-z+1\right)^2+y^2+2z^2-4y+2yz-6z+2009$

$=\left[x-\left(y-z+1\right)\right]^2+y^2-2y\left(2-z\right)+\left(2-z\right)^2-\left(2-z\right)^2+2z^2-6z+2009$

$=\left(x-y+z-1\right)^2+\left(y-2+z\right)^2+z^2-2z+2005$

$=\left(x-y+z-1\right)^2+\left(y-2+z\right)^2+\left(z-1\right)^2+2004\ge2004$

Dấu "=" xảy ra $\Leftrightarrow\left\{{}\begin{matrix}x-y+z-1=0\y-2+z=0\z-1=0\end{matrix}\right.$ $\Leftrightarrow x=y=z=1$

Vậy $B_{min}=2004\Leftrightarrow x=y=z=1$Câu trả lời: $A=x^2+2y^2+3z^2-2xy+2xz-2x-2y-8z+2010$

$=x^2-2x\left(y-z+1\right)+\left(y-z+1\right)^2+y^2+2z^2-4y+2yz-6z+2009$

$=\left[x-\left(y-z+1\right)\right]^2+y^2-2y\left(2-z\right)+\left(2-z\right)^2-\left(2-z\right)^2+2z^2-6z+2009$

$=\left(x-y+z-1\right)^2+\left(y-2+z\right)^2+z^2-2z+2005$

$=\left(x-y+z-1\right)^2+\left(y-2+z\right)^2+\left(z-1\right)^2+2004\ge2004$

Dấu "=" xảy ra $\Leftrightarrow\left\{{}\begin{matrix}x-y+z-1=0\y-2+z=0\z-1=0\end{matrix}\right.$ $\Leftrightarrow x=y=z=1$

Vậy $B_{min}=2004\Leftrightarrow x=y=z=1$

Anh Nguyễn Quang · Answer

Thanks for answering!!!!!Câu trả lời: Thanks for answering!!!!!