Ta có : \(\frac{x}{y}=\frac{y}{z}=\frac{z}{x}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(\frac{x}{y}=\frac{y}{z}=\frac{z}{x}=\frac{x+y+z}{y+z+x}=1\) (do \(x+y+z\ne0\))

\(\Leftrightarrow\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}}\)

\(\Leftrightarrow x=y=z\)

Thay \(x=y=z\) vào \(N=\frac{x^{123}.y^{456}}{z^{579}}\), ta có :

\(N=\frac{x^{123}.x^{456}}{x^{579}}\)

\(\Leftrightarrow\frac{x^{579}}{x^{579}}=1\)

Vậy N = 1

fai fai ối dồi ôi luôn

\(x+y+z=0\)

⇔\(-x=y+z\)

⇔\(x^2=\left(y+z\right)^2\) 

⇔\(x^2=y^2+2yz+z^2\) 

⇔\(y^2+z^2-x^2=-2yz\)

Tương tự:

\(z^2+x^2-y^2=-2zx\)

\(x^2+y^2-z^2=-2xy\)

➞ S = \(\dfrac{1}{-2xy}+\dfrac{1}{-2yz}+\dfrac{1}{-2zx}=\dfrac{x+y+z}{-2xyz}=0\) 

Vậy S = 0

Ta có:

\(x+y+z=0\)

\(\Rightarrow\left(x+y\right)^2=\left(-z\right)^2\)

\(\Rightarrow x^2+y^2+2xy=z^2\)

\(\Rightarrow x^2+y^2-z^2=-2xy\)

Tương tự ta được:
\(S=\frac{1}{-2xy}+\frac{1}{-2yz}+\frac{1}{-2zx}=-\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=-\frac{1}{2}\cdot\frac{x+y+z}{xyz}=0\)

Vậy S=0

\(\frac{1}{x}+\frac{1}{y}=2-\frac{1}{z}\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}=4+\frac{1}{z^2}-\frac{4}{z}\)

\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}=-\frac{4}{z}\) \(\Rightarrow\frac{1}{z}=-\frac{1}{4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\)

\(\Rightarrow\frac{1}{x}+\frac{1}{y}-\frac{1}{4}\left(\frac{1}{x^2}+\frac{1}{y^2}\right)=2\Rightarrow\frac{1}{4x^2}-\frac{1}{x}+1+\frac{1}{4y^2}-\frac{1}{y}+1=0\)

\(\Rightarrow\left(\frac{1}{2x}-1\right)^2+\left(\frac{1}{2y}-1\right)^2=0\Rightarrow\left\{{}\begin{matrix}\frac{1}{2x}-1=0\\\frac{1}{2y}-1=0\end{matrix}\right.\)

\(\Rightarrow x=y=\frac{1}{2}\Rightarrow\frac{1}{z}=2-\left(\frac{1}{x}+\frac{1}{y}\right)=-2\Rightarrow z=-\frac{1}{2}\)

\(\Rightarrow P=\left(\frac{1}{2}+1-\frac{1}{2}\right)^{2018}=1^{2018}=1\)