Tìm x, y, z biết (y+z+1)/x=(x+z+2)/y=(x+y-3)/z=1/(x+y+z)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

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

Theo đề bài ta có:

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

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

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

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

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

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

. \(x+y+z=\dfrac{1}{2}\Rightarrow y+z=\dfrac{1}{2}-x\)

Thay \(y+z+1=2x\) ta được \(\dfrac{1}{2}-x+1=2x\)

\(\Rightarrow\dfrac{3}{2}=3x\)

\(\Rightarrow x=\dfrac{1}{2}\)

. \(x+y+z=\dfrac{1}{2}\Rightarrow x+z=\dfrac{1}{2}-y\)

Thay \(x+z+2=2y\) ta được \(\dfrac{1}{2}-y+2=2y\)

\(\Rightarrow\dfrac{5}{2}=3y\)

\(\Rightarrow y=\dfrac{5}{6}\)

\(\Rightarrow x+y+z=\dfrac{1}{2}+\dfrac{5}{6}+z=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{4}{3}+z=\dfrac{1}{2}\)

\(\Rightarrow z=\dfrac{1}{2}-\dfrac{4}{3}\)

\(\Rightarrow z=-\dfrac{5}{6}\)

Vậy \(x=\dfrac{1}{2}\) , \(y=\dfrac{5}{6}\) , \(z=-\dfrac{5}{6}\)