Question

Tìm các số x;y;z biết

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) Và x-2y+3z=14

Answer 1

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=>\frac{x-1}{2}=\frac{2\left(y-2\right)}{6}=\frac{3\left(z-3\right)}{12}=>\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}\)

Theo t/c dãy tỉ số=nhau:

\(\frac{x-1}{2}=\frac{2y-4}{6}=\frac{3z-9}{12}=\frac{x-1-\left(2y-4\right)+\left(3z-9\right)}{2-6+12}=\frac{x-1-2y+4+3z-9}{8}\)

\(=\frac{\left(x-2y+3z\right)-\left(1-4+9\right)}{8}=\frac{14-6}{8}=\frac{8}{8}=1\)

Do đó: \(\frac{x-1}{2}=1=>x-1=2=>x=3\)

\(\frac{y-2}{3}=1=>y-2=3=>y=5\)

\(\frac{z-3}{4}=1=>z-3=4=>z=7\)

Vậy x=3;y=5;z=7