Mình nghĩ phải là \(\frac{z}{5}\)mới hợp lý, nếu có gì sai thì bạn comment nhé
Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=k\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\)
Khi đó : \(\left(3k\right)^2+2\cdot\left(4k\right)^2+4\cdot\left(5k\right)^2=141\)
\(\Leftrightarrow141k^2=141\)
\(\Leftrightarrow k^2=1\)
\(\Leftrightarrow k=\pm1\)
TH1 : \(\left\{{}\begin{matrix}x=3\\y=4\\z=5\end{matrix}\right.\)
TH2 : \(\left\{{}\begin{matrix}x=-3\\y=-4\\z=-5\end{matrix}\right.\)
Vậy....
\(\frac{x}{3}=\frac{y}{4}=\frac{z}{3}\Rightarrow\frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{9}=\frac{x^2}{9}=\frac{2y^2}{32}=\frac{4z^2}{36}\)
\(=\frac{x^2+2y^2+4z^2}{9+32+36}=\frac{141}{77}\) ( theo tính chất dãy tỉ số bằng nhau )
\(\Rightarrow\left\{{}\begin{matrix}x^2=\frac{141\cdot9}{77}=\frac{1269}{77}\\y^2=\frac{141\cdot16}{77}=\frac{2256}{77}\\z^2=\frac{1269}{77}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm\frac{3\sqrt{141}}{\sqrt{77}}\\y=\pm\frac{4\sqrt{141}}{\sqrt{77}}\\z=\pm\frac{3\sqrt{141}}{\sqrt{77}}\end{matrix}\right.\)
+ \(\frac{x}{3}=\frac{y}{4}=\frac{z}{3}\) => x,y,z cùng dấu
\(\Rightarrow\left\{{}\begin{matrix}x=z=\frac{3\sqrt{141}}{\sqrt{77}}\\y=\frac{4\sqrt{141}}{\sqrt{77}}\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=z=\frac{-3\sqrt{141}}{\sqrt{77}}\\y=\frac{-4\sqrt{141}}{\sqrt{77}}\end{matrix}\right.\)
