Ta có: \(\frac{x}{y}=\frac{2}{3}\)
=> \(\frac{x}{2}=\frac{y}{3}\)=> \(\frac{x}{6}=\frac{y}{9}\)(1)
Có: \(\frac{x}{3}=\frac{z}{5}\)=> \(\frac{x}{6}=\frac{z}{10}\)(2)
Từ (1) ; (2) => \(\frac{x}{6}=\frac{y}{9}=\frac{z}{10}\)=> \(\frac{x^2}{36}=\frac{y^2}{81}=\frac{z^2}{100}=\frac{x^2+y^2+z^2}{36+81+100}=\frac{\frac{217}{4}}{217}=\frac{1}{4}\)
=> \(\hept{\begin{cases}\frac{x^2}{36}=\frac{1}{4}\\\frac{y^2}{81}=\frac{1}{4}\\\frac{z^2}{100}=\frac{1}{4}\end{cases}}\)=> \(\hept{\begin{cases}x^2=9\\y^2=\frac{81}{4}\\z^2=25\end{cases}}\)
Vì x, y, z dương nên suy ra: \(\hept{\begin{cases}x=3\\y=\frac{9}{2}\\z=5\end{cases}}\)
=> \(x+2y-2z=3+2.\frac{9}{2}-2.5=2\)
Ta có : \(\frac{x}{y}=\frac{2}{3};\frac{x}{3}=\frac{z}{5}\)
\(\Rightarrow\frac{x}{2}=\frac{y}{3};\frac{x}{3}=\frac{z}{5}\)
\(\Rightarrow\frac{x}{6}=\frac{y}{9};\frac{x}{6}=\frac{z}{10}\)
\(\Rightarrow\frac{x}{6}=\frac{y}{9}=\frac{z}{10}\)
Đặt \(\frac{x}{6}=\frac{y}{9}=\frac{z}{10}=k\)(k>0)
\(\Rightarrow\hept{\begin{cases}x=6k\\y=9k\\z=10k\end{cases}}\)
Thay x=6k; y=9k; z=10k vào \(x^2+y^2+z^2=\frac{217}{4}\) ta có:
\(\left(6k\right)^2+\left(9k\right)^2+\left(10k^2\right)=\frac{217}{4}\)
\(\Rightarrow6^2.k^2+9^2.k^2+10^2.k^2=\frac{217}{4}\)
\(\Rightarrow k^2.\left(6^2+9^2+10^2\right)=\frac{217}{4}\)
\(\Rightarrow k^2.\left(36+81+100\right)=\frac{217}{4}\)
\(\Rightarrow k^2.217=\frac{217}{4}\)
\(\Rightarrow k^2=\frac{217}{4}.\frac{1}{217}=\frac{1}{4}\)
\(\Rightarrow k=\pm\frac{1}{2}\)
Mà k >0
\(\Rightarrow k=\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}x=6.\frac{1}{2}=3\\y=9.\frac{1}{2}=\frac{9}{2}\\z=10.\frac{1}{2}=5\end{cases}}\)( thỏa mãn x;y dương)
\(\Rightarrow x+2y-2z=3+2.\frac{9}{2}-2.5=3+9-10=2\)
Vậy x+2y-2z=2
