$(3x+6)^2+5+|3-y|+|x+z-y|≤5-|3y-9|$
$↔(3x+6)^2+5+|3-y|+|x+z-y|=5-3|y-3|$
$↔(3x+6)^2+|3-y|+3|y-3|+|x+z-y|≤0$
$↔(3x+6)^2+4|y-3|+|x+z-y|≤0$
Vì: $\begin{cases}(3x+6)^2≥0\\4|y-3|≥0\\|x+z-y|≥0\end{cases}$
$→(3x+6)^2+4|y-3|+|x+z-y|≥0$
Mà: $(3x+6)^2+4|y-3|+|x+z-y|≤0$
$→\begin{cases}3x+6=0\\4(y-3)=0\\x+z-y=0\end{cases}$
$→\begin{cases}x=-2\\y=3\\z=5\end{cases}$
Vậy: $(x,y,z)=(-2;3;5)$
