x+y+z=1?

Giải:

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

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

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

Xét \(\frac{x}{y+z+1}=\frac{1}{2}\)

\(\Rightarrow2x=y+z+1\)

\(\Rightarrow3x=x+y+z+1\)

\(\Rightarrow3x=1+1\)

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

Xét \(\frac{y}{x+z+1}=\frac{1}{2}\)

\(\Rightarrow2y=x+z+1\)

\(\Rightarrow3y=x+y+z+1\)

\(\Rightarrow3y=1+1\)

\(\Rightarrow y=\frac{2}{3}\)

Xét \(\frac{z}{x+y-2}=\frac{1}{2}\)

\(\Rightarrow2z=x+y-2\)

\(\Rightarrow3z=x+y+z-2\)

\(\Rightarrow3z=1-2\)

\(\Rightarrow z=\frac{-1}{3}\)

Từ đó \(\frac{x+y}{z+1}=\frac{\frac{2}{3}+\frac{2}{3}}{\frac{-1}{3}+1}=\frac{\frac{4}{3}}{\frac{2}{3}}=\frac{4}{2}=2\)

Vậy \(\frac{x+y}{z+1}=2\)

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

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

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

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