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Ta có \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xyz}\left(x+y+z\right)=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{xyz}=4\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)(vì \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}>0\))
Mặt khác, ta có : \(\frac{1}{x+y+z}=2\) .
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\left(\frac{1}{z}-\frac{1}{x+y+z}\right)=0\)
\(\Leftrightarrow\frac{x+y}{xy}+\frac{x+y}{z\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(\frac{1}{xy}+\frac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Leftrightarrow\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz\left(x+y+z\right)}=0\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
=> x+y = 0 hoặc y + z = 0 hoặc z + x = 0
Từ đó suy ra P = 0 (lí do vì x,y,z là các số mũ lẻ)
+ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\Rightarrow\frac{1}{z}=2-\frac{1}{x}-\frac{1}{y}\)
\(\Rightarrow\frac{1}{z^2}=\left(2-\frac{1}{x}-\frac{1}{y}\right)^2\)
+ \(\frac{2}{xy}-\frac{1}{z^2}=4\Rightarrow\frac{2}{xy}-\left(2-\frac{1}{x}-\frac{1}{y}\right)^2=4\)
\(\Rightarrow\frac{2}{xy}-\left(4+\frac{1}{x^2}+\frac{1}{y^2}-\frac{4}{x}-\frac{4}{y}+\frac{2}{xy}\right)=4\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}-\frac{4}{x}-\frac{4}{y}+8=0\)
\(\Rightarrow\left(\frac{1}{x}-2\right)^2+\left(\frac{1}{y}-2\right)^2=0\) \(\Rightarrow\left\{{}\begin{matrix}\left(\frac{1}{x}-2\right)^2=0\\\left(\frac{1}{y}-2\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}=2\\\frac{1}{y}=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{2}\\\frac{1}{z}=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{2}\\z=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow P=\left(\frac{1}{2}+1-\frac{1}{2}\right)^{2018}=1\)