Giải:
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\)
\(\Rightarrow x=2k,y=3k,z=5k\)
Ta có: \(x^2-2y^2+z^2=44\)
\(\Rightarrow\left(2k\right)^2-2\left(3k\right)^2+\left(5k\right)^2=44\)
\(\Rightarrow4.k^2-18.k^2+25.k^2=44\)
\(\Rightarrow\left(4-18+25\right)k^2=44\)
\(\Rightarrow11k^2=44\)
\(\Rightarrow k^2=4\)
\(\Rightarrow k=\pm2\)
+) \(k=2\Rightarrow x=4;y=6;z=10\)
\(\Rightarrow\left(x+y+z\right)^2=400\)
+) \(k=-2\Rightarrow x=-4;y=-6;z=-10\)
\(\Rightarrow\left(x+y+z\right)^2=400\)
Vậy \(\left(x+y+z\right)^2=400\)
\(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{25}\Rightarrow\frac{x^2}{4}=\frac{2y^2}{18}=\frac{z^2}{25}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x^2}{4}=\frac{2y^2}{18}=\frac{z^2}{25}=\frac{x^2-2y^2+z^2}{4-18+25}=\frac{44}{11}=4\)
\(\Rightarrow\left\{\begin{matrix}\frac{x}{2}=4\rightarrow x=4\cdot2=8\\\frac{y}{3}=4\rightarrow y=4\cdot3=12\\\frac{z}{5}=4\rightarrow z=4\cdot5=20\end{matrix}\right.\)
\(\Rightarrow\left(x+y+z\right)^2=\left(8+12+20\right)^2=40^2=1600\)