+) Nếu \(x+y+z\ne0\)
Áp dụng t/c dãy tỉ số = nhau , ta có :
\(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}\)
\(\Rightarrow x+y+z=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{y+z+1}=\dfrac{1}{2}\\\dfrac{y}{x+z+1}=\dfrac{1}{2}\\\dfrac{z}{x+y-2}=\dfrac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x=y+z+1\\2y=x+z+1\\2z=x+y-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x=x+y+z+1\\3y=x+y+z+1\\3z=x+y+z-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3x=\dfrac{1}{2}+1\\3y=\dfrac{1}{2}+1\\3z=\dfrac{1}{2}-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x=\dfrac{3}{2}\\3y=\dfrac{3}{2}\\3z=\dfrac{-3}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{2}\\z=\dfrac{-1}{2}\end{matrix}\right.\)
+) Nếu x + y + z = 0
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{y+z+1}=0\\\dfrac{y}{x+z+1}=0\\\dfrac{z}{x+y-2}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\y=0\\z=0\end{matrix}\right.\)
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