\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{x}=\dfrac{x+y+z}{y+z+x}=1\)
\(\Rightarrow x=y=z\)
\(\Rightarrow P=\dfrac{x^{20}.z^{20}}{y^{50}}=\dfrac{x^{20}.x^{20}}{x^{50}}=\dfrac{x^{40}}{x^{50}}=\dfrac{1}{x^{10}}\)
Ta có :
\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{x}\) và \(x+y+z\ne0\)\(\Rightarrow\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{x}=\dfrac{x+y+z}{y+z+x}=1\)
*\(\dfrac{x}{y}=1\Rightarrow x=y\left(1\right)\)
*\(\dfrac{y}{z}=1\Rightarrow y=z\left(2\right)\)
Từ (1) và (2) suy ra : x = y = z
Ta lại có :
\(P=\dfrac{x^{20}.z^{20}}{y^{50}}=\dfrac{x^{20}.x^{20}}{x^{50}}=\dfrac{y^{20}.y^{20}}{y^{50}}=\dfrac{z^{20}.z^{20}}{z^{50}}=\dfrac{x^{40}}{x^{50}}=\dfrac{y^{40}}{y^{50}}=\dfrac{z^{40}}{z^{50}}=x^{-10}=y^{-10}=z^{-10}\)
\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{x}\)
Áp dụng tính chất dãy tỉ số bang nhau ta có :
\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{x}=\dfrac{x+y+z}{y+z+x}=1\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \dfrac{x}{y}=1\Rightarrow x=y\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\\)
\(\dfrac{y}{z}=1\Rightarrow y=z\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\\)
Do do:x=y=z
P\(\dfrac{x^{20}.z^{20}}{y^{50}}=\dfrac{x^{20}.x^{20}}{x^{50}}=\dfrac{x^{40}}{x^{50}}=\dfrac{1}{x^{10}}\)
Vậy P=\(\dfrac{1}{x^{10}}\)
