Question

Tìm các số 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

\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)(đk x+y+z\(\ne0\)

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

\(\Rightarrow\frac{1}{x+y+z}=2\Rightarrow x+y+z=0,5\)

\(\Rightarrow y+z=0,5-x,x+z=0,5-y,x+y=0,5-z\)

\(\Rightarrow\frac{0,5-x+1}{x}=2\Rightarrow\frac{1,5-x}{x}=2\Rightarrow1,5-x=2x\Rightarrow3x=1,5\Rightarrow x=\frac{1}{2}\)

\(\Rightarrow\frac{0,5-y+2}{y}=2\Rightarrow\frac{2,5-y}{y}=2\Rightarrow2,5-y=2y\Rightarrow3y=2,5\Rightarrow y=\frac{5}{6}\)

\(\Rightarrow z=0,5-\frac{1}{2}-\frac{5}{6}=-\frac{5}{6}\)

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