ta có \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=>\dfrac{x^2}{2}=\dfrac{y^2}{3}=\dfrac{z^2}{4}\) \(=>\dfrac{x}{4}=\dfrac{y}{9}=\dfrac{z}{16}\) và \(x^2+y^2+z^2=29\)
ADTC dãy tỉ số bằng nhau ta có
\(\dfrac{x}{4}=\dfrac{y}{9}=\dfrac{z}{16}=\dfrac{x^2+y^2+z^2}{4+9+16}=\dfrac{29}{29}=1\)
\(=>\left\{{}\begin{matrix}x=1.4=4\\y=1.9=9\\z=1.16=16\end{matrix}\right.\)