Áp dụng tính chất dãy tỉ số bằng nhau
$⇒\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{7}=\dfrac{x+y+z}{3+4+7}=\dfrac{40,5}{14}=\dfrac{81}{28}$
\(⇒\left[ \begin{array}{l}\dfrac{x}{3}=\dfrac{81}{28}⇒x=\dfrac{243}{28}\\\dfrac{y}{4}=\dfrac{81}{28}⇒y=\dfrac{81}{7}\\\dfrac{z}{7}=\dfrac{81}{28}⇒z=\dfrac{81}{4}\end{array} \right.\)
Vậy $(x,y,z)=(\dfrac{243}{28},\dfrac{81}{7},\dfrac{81}{4})$