Tìm Min A=\(x^2+2y^2+3z^2-2xy+2xz-2x-2y-8z+2008\)
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\(A=x^2+2y^2+3z^2-2xy+2xz-2x-2y-8z+2010\)
\(=x^2-2x\left(y-z+1\right)+\left(y-z+1\right)^2+y^2+2z^2-4y+2yz-6z+2009\)
\(=\left[x-\left(y-z+1\right)\right]^2+y^2-2y\left(2-z\right)+\left(2-z\right)^2-\left(2-z\right)^2+2z^2-6z+2009\)
\(=\left(x-y+z-1\right)^2+\left(y-2+z\right)^2+z^2-2z+2005\)
\(=\left(x-y+z-1\right)^2+\left(y-2+z\right)^2+\left(z-1\right)^2+2004\ge2004\)
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 \(B_{min}=2004\Leftrightarrow x=y=z=1\)
\(D=x^2+y^2+z^2-2xy+2zx-2yz+y^2+2z^2+2yz-2\left(x-y+z\right)-4y-6z+19\)
\(=\left(x-y+z\right)^2-2\left(x-y+z\right)+1+\left(y^2+z^2+2yz-4y-4z+4\right)+z^2-2z+1+13\)
\(=\left(x-y+z-1\right)^2+\left(y+z-2\right)^2+\left(z-1\right)^2+13\ge13\)
\(D_{min}=13\) khi \(\left\{{}\begin{matrix}x-y+z=1\\y+z=2\\z=1\end{matrix}\right.\) \(\Rightarrow x=y=z=1\)
A=\(x^2+2y^2+3z^2-2xy+2xz-2x-2y-8z+2008\)
A=\(\left(x^2+y^2+z^2+1-2xy+2xz-2x+2y-2z\right)+\left(y^2-4y+4\right)+2\left(z^2-2.\frac{3}{2}z+\frac{9}{4}\right)+1998,5\)A=\(\left(x-y+z-1\right)^2+\left(y-2\right)^2+2\left(z-\frac{3}{2}\right)^2+1998,5\)
vậy A min = 1998,5↔\(\begin{cases}x-y+z-1=0\\y-2=0\\z-\frac{3}{2}=0\end{cases}\)↔\(\begin{cases}x=z=1,5\\y=2\end{cases}\)
(thế wai nào thử lại vẫn sai z,thánh nào cao tay jup vs)