Question

Answer 1

\(\dfrac{y+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{2\left(x+y+z\right)}{x+y+z}=2\)(t/c dãy tỉ số bằng nhau)

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

Ta có: \(\dfrac{\dfrac{1}{2}-x+1}{x}=2\Rightarrow x=\dfrac{1}{2}\)

\(\Rightarrow y+z=0\Rightarrow y=-z\)

Ta có:

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