Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{x}{3}=\frac{y}{8}=\frac{z}{5}\Rightarrow\frac{4x}{12}=\frac{3y}{24}=\frac{2z}{10}=\frac{4x+4y-2z}{12+24-10}=\frac{96}{26}=\frac{48}{13}\)
\(\Rightarrow x=\frac{48}{13}\times3=\frac{144}{13}\)
\(y=\frac{48}{13}\times8=\frac{384}{13}\)
\(z=\frac{48}{13}\times5=\frac{240}{13}\)
Vậy ....
Áp dụng t/c DTSBN ta có:
\(\frac{x}{3}=\frac{y}{8}=\frac{z}{5}=\frac{4x+3y-2z}{4.3+3.8-2.5}=\frac{48}{13}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{3}=\frac{48}{13}\Rightarrow x=\frac{144}{13}\\\frac{y}{8}=\frac{48}{13}\Rightarrow y=\frac{384}{13}\\\frac{z}{5}=\frac{48}{13}\Rightarrow z=\frac{240}{13}\end{cases}}\)
Vậy \(x=\frac{144}{13};y=\frac{384}{13};z=\frac{240}{13}\)
hok tốt!