Question

Biết \(\left(x_0+y_0+z_0\right)\) là nghiệm nguyên dương của phương trình:

\(x^2+y^2+z^2=xy+3y+2z-4\)

Tính \(x_0+y_0+z_0\)

Answer 1

\(\Leftrightarrow x^2+y^2+z^2-xy-3y-2z=-4\)

\(\Leftrightarrow\left(x^2+\dfrac{y^2}{4}-xy\right)+\dfrac{3}{4}\left(y^2-4y+4\right)+\left(z^2-2z+1\right)=-4+4=0\)

\(\Leftrightarrow\left(x-\dfrac{y}{2}\right)^2+\dfrac{3}{4}\left(y-2\right)^2+\left(z-1\right)^2=0\)

\(\left\{{}\begin{matrix}z_o-1=0\\y_o-2=0\\x_o-\dfrac{y_o}{2}=0\\\end{matrix}\right.\) \(\left\{{}\begin{matrix}2z_o=2\\3y_o=6\\2x_o-y_o=0\\2\left(x_o+y_o+z_o\right)=8\end{matrix}\right.\) \(\Rightarrow x_o+y_o+z_o=4\)

Answer 2

ta có: \(x^2+y^2+z^2-xy-3y-2z+4=0\)

\(\left(x^2-xy+\dfrac{1}{4}y^2\right)+\left(\dfrac{3}{4}y^2-3y+3\right)+\left(z^2-2z+1\right)=0\)

\((x-\dfrac{1}{2}y)^2+3\left(\dfrac{1}{2}y-1\right)^2+\left(z-1\right)^2=0\)

giải 3 bình phương để bằng 0 được x=1;y=2;z=1

Answer 3

\(x_0+y_0+z_0=1+2+1=4\)