TH1:x+y+z≠0
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{y+z}=\dfrac{y}{x+z}=\dfrac{z}{x+y}=\dfrac{x+y+z}{y+z+x+z+x+y}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}\)
\(\dfrac{x}{y+z}=\dfrac{1}{2}\Rightarrow2x=y+z\Rightarrow3x=x+y+z\left(1\right)\\ \dfrac{y}{x+z}=\dfrac{1}{2}\Rightarrow2y=x+z\Rightarrow3y=x+y+z\left(2\right)\\ \dfrac{z}{x+y}=\dfrac{1}{2}\Rightarrow2z=x+y\Rightarrow3z=x+y+z\left(3\right)\)
Từ (1), (2), (3) \(\Rightarrow3x=3y=3z\Rightarrow x=y=z\)
\(Q=\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}\\ =\dfrac{x+x}{x}+\dfrac{x+x}{x}+\dfrac{x+x}{x}\\ =\dfrac{2x}{x}+\dfrac{2x}{x}+\dfrac{2x}{x}\\ =2+2+2\\ =6\)
TH2:x+y+z=0
\(\Rightarrow\left\{{}\begin{matrix}x+y=-z\\y+z=-x\\x+z=-y\end{matrix}\right.\)
\(Q=\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}\\ =\dfrac{-x}{x}+\dfrac{-y}{y}+\dfrac{-z}{z}\\ =\left(-1\right)+\left(-1\right)+\left(-1\right)\\ =-3\)
