a,(x-4)²-(y-1)²

k cho mk nhe

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)

\(a,4x^2+9y^2+4x-24y+17=0\)

\(\Rightarrow\left(4x^2+4x+1\right)+\left(9y^2-24y+16\right)=0\)

\(\Rightarrow\left(2x+1\right)^2+\left(3y-4\right)^2=0\)

\(\left(2x+1\right)^2\ge0;\left(3y-4\right)^2\ge0\)

\(\Rightarrow\hept{\begin{cases}\left(2x+1\right)^2=0\\\left(3y-4\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x+1=0\\3y-4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{4}{3}\end{cases}}}\)

\(A=x^2+2y^2+3z^2-2xy+2xz-2x-2y-8z+2010\)

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

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

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

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

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 \(B_{min}=2004\Leftrightarrow x=y=z=1\)