Question

Tìm x; y ; z biết: \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)

Answer 1

\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=\frac{1}{x+y+z}=2\)

\(\Rightarrow x+y+z=\frac{1}{2}\)

\(y+z+1=2x\Rightarrow x+y+z+1=3x\Rightarrow3x=1+\frac{1}{2}=\frac{3}{2}\Rightarrow x=\frac{1}{2}\)

\(x+z+2=2y\Rightarrow x+y+z+2=3y\Rightarrow3y=2+\frac{1}{2}=\frac{5}{2}\Rightarrow y=\frac{5}{6}\)

\(x+y-3=2z\Rightarrow x+y+z-3=3z\Rightarrow3z=\frac{1}{2}-3=-\frac{5}{2}\Rightarrow z=-\frac{5}{6}\)