ta có: \(\frac{x-1}{5}\) = \(\frac{y-2}{3}\) = \(\frac{z-2}{2}\) => \(\frac{3x-3}{15}=\frac{5y-10}{15}=\frac{6z-12}{12}\) và 3x-5y+6z =9
Áp dụng t/c ..., ta có:
\(\frac{3x-3}{15}=\frac{5y-10}{15}=\frac{6z-12}{12}\) =\(\frac{\left(3x-5y+6z\right)+\left(-3+10-12\right)}{15-15+12}\) =\(\frac{4}{12}\)=\(\frac{1}{3}\)
\(\frac{x-1}{5}\) =\(\frac{1}{3}\) =>x-1=\(\frac{5}{3}\)=>x=\(\frac{8}{3}\)
\(\frac{y-2}{3}\) = \(\frac{1}{3}\)=>y-2=1 =>y=3
\(\frac{z-2}{2}\) =\(\frac{1}{3}\) =>z-2=\(\frac{2}{3}\) =>z=\(\frac{8}{3}\)