Sửa đề:
\(\dfrac{x}{x+y+1}=\dfrac{y}{x+z+1}=\dfrac{z}{z+y-2}\)
Dựa vào t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{x+y+1}=\dfrac{y}{x+z+1}=\dfrac{z}{z+y-2}=\dfrac{x+y+z}{x+y+x+z+z+y+\left(1+1-2\right)}=\dfrac{x+y+z}{x+x+y+y+z+z}=\dfrac{1\left(x+y+z\right)}{2\left(x+y+z\right)}=\dfrac{1}{2}\)\(x+y+z=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{y}{x+z+1}=\dfrac{1}{2}\)
\(2y=x+z+1\)
\(3y=\dfrac{1}{2}+1\)
\(y=\dfrac{1}{2}\)
Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\dfrac{x}{x+y+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}=x+y+z\)
\(\Rightarrow\dfrac{y}{x+z+1}=\dfrac{1}{2}\)
\(\Rightarrow2y=x+z+1\)
\(\Rightarrow3y=x+y+z+1\)
\(\Rightarrow3y=\dfrac{1}{2}+1\)
\(\Rightarrow y=\dfrac{1}{2}\)
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