Giải:
Ta có: \(x^2=yz\Rightarrow\dfrac{x}{y}=\dfrac{z}{x}\)
\(y^2=xz\Rightarrow\dfrac{x}{y}=\dfrac{y}{z}\)
\(z^2=xy\Rightarrow\dfrac{z}{x}=\dfrac{y}{z}\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{x}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{x}=\dfrac{x+y+z}{x+y+z}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{y}=1\Rightarrow x=y\\\dfrac{y}{z}=1\Rightarrow y=z\\\dfrac{z}{x}=1\Rightarrow z=x\end{matrix}\right.\Rightarrow x=y=z\left(đpcm\right)\)
Vậy x = y = z