Question

Tim ba so x, y, z biet \(\frac{y+z+1}{x}=\frac{x+ z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)

Answer 1

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

Suy ra

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

\(y+z+1=2x\)(2)

\(x+z+2=2y\)(3)

\(x+y-3=2z\)(4)

(2)-(1) ta có

\(1-x=2x-\frac{1}{2}\Rightarrow3x=\frac{3}{2}\Rightarrow x=\frac{1}{2}\)

\(x+y+z=\frac{1}{2}\Rightarrow y+z=\frac{1}{2}-x\Leftrightarrow y+z=\frac{1}{2}-\frac{1}{2}=0\)

\(y=-z\)

\(x+z+2=\frac{1}{2}+2-y==\frac{5}{2}-y\)

\(\frac{\frac{5}{2}-y}{y}=\frac{5}{2y}-1=2\Leftrightarrow\frac{5}{2y}=3\Leftrightarrow y=\frac{5}{6}\)

\(z=-\frac{5}{6}\)