+) Với \(x+y+z=0\)
\(\Leftrightarrow\dfrac{x}{z+y+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=0\)
\(\Leftrightarrow x=y=z=0\left(loại\right)\)
+) Với \(x+y+z\ne0\)
Theo t,c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{z+y+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=x+y+z=\dfrac{x+y+z}{2\left(x+y+z\right)+\left(1+1-2\right)}=\dfrac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{z+y+1}=\dfrac{1}{2}\\\dfrac{y}{x+z+1}=\dfrac{1}{2}\\\dfrac{z}{x+y-2=\dfrac{1}{2}}\end{matrix}\right.\) Và \(x+y+z=\dfrac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}z+y+1=2x\\x+z+1=2y\\x+y-2=2z\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z+1=3x\\x+y+z+1=3y\\x+y+z-2=3z\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}+1=3x\\\dfrac{1}{2}+1=3y\\\dfrac{1}{2}-2=3z\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2}=3x\\\dfrac{3}{2}=3y\\-\dfrac{3}{2}=3z\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{2}\\z=-\dfrac{1}{2}\end{matrix}\right.\)
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