Ta có : \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\Rightarrow\frac{x^2}{4}=\frac{3y^2}{27}=\frac{2z^2}{32}\)
Theo tính chất dãy tỉ số bằng nhau ta có :
\(\frac{x^2}{4}=\frac{3y^2}{27}=\frac{2z^2}{32}=\frac{x^2+3y^2-2z^2}{4+27-32}=\frac{-16}{-1}=16\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=16.4\\3y^2=16.27\\2z^2=16.32\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=64\\y^2=144\\z^2=256\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\pm8\\y=\pm12\\z=\pm16\end{matrix}\right.\)
Vậy : \(\left(x,y,z\right)\in\left\{\left(8,12,16\right);\left(-8,-12,-16\right)\right\}\)
Chúc bạn học tốt !
\(\frac{x}{2}\)=\(\frac{y}{3}=\frac{z}{4}\)
=>\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{16}\)
=>\(\frac{x^2}{9}=\frac{3y^2}{27}=\frac{2z^2}{32}=\frac{x^2+3y^2-2z^2}{9+27-32}=\frac{16}{4}=4\)
=>\(x^2=36\)
=>x=6 hoặc x=-6
x=6=>y=9;z=12
x=-6=>y=-9;z=-12
