\(3x=5y=4z\Leftrightarrow\frac{x}{20}=\frac{y}{12}=\frac{z}{15}=\frac{5x}{100}=\frac{4y}{48}\)
Áp dụng t/c dãy tỉ số bằng nhau: \(\frac{x}{20}=\frac{y}{12}=\frac{z}{15}=\frac{5x}{100}=\frac{4y}{48}=\frac{5x-4y+z}{100-48+15}=\frac{-1}{67}\)
=>\(x=\frac{-1}{67}.20=-\frac{20}{67};y=\frac{-1}{67}.12=-\frac{12}{67};z=\frac{-1}{67}.15=-\frac{15}{67}\)
Vậy ......