\(AD...,có:\dfrac{x+2}{7}=\dfrac{y-3}{5}=\dfrac{z}{3}=\dfrac{x+2+y-3-z}{7+5-3}=\dfrac{x+y-z-1}{9}=\dfrac{-17-1}{9}=-2\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x+2}{7}=-2\\\dfrac{y-3}{5}=-2\\\dfrac{z}{3}=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-16\\y=-7\\z=-6\end{matrix}\right.\)
\(\dfrac{\left(x+2\right)}{7}=\dfrac{\left(y-3\right)}{5}=\dfrac{z}{3}=\dfrac{\left(x+2+y-3-z\right)}{7+5-3}=\dfrac{\left(x+y-z-1\right)}{9}=-\dfrac{18}{9}=-2\)
\(=>\dfrac{\left(x+2\right)}{7}=-2\)\(=>x=-2.7-2=-16\)
\(=>\dfrac{\left(y-3\right)}{5}=-2=>y=-2.5+3=-7\)\(;\dfrac{z}{3}=-2=>z=-2.3=-6\)
