\(\dfrac{x}{x+y+z}=\dfrac{y}{x+z+t}=\dfrac{z}{y+z+t}=\dfrac{t}{x+z+t}\\ =\dfrac{x+y+z+t}{x+y+z+x+z+y+y+z+t+x+z+t}\)

\(=\dfrac{x+y+z+t}{3\left(x+y+z+t\right)}=\dfrac{1}{3}\\ hayM=\dfrac{1}{3}\)

\(M^{10}=\left(\dfrac{1}{3}\right)^{10}=\dfrac{1}{3^{10}}< 2017\)

\(M>\dfrac{x}{x+y+z+t}+\dfrac{y}{x+y+z+t}+\dfrac{z}{x+y+z+t}+\dfrac{t}{x+y+z+t}=1\)

\(\dfrac{a}{b}< 1\Rightarrow\) \(\dfrac{a}{b}< \dfrac{a+m}{b+m}\) (Bạn chứng minh qua nhân chéo nhé)

\(\Rightarrow M< \dfrac{x+t}{x+y+z+t}+\dfrac{y+z}{x+y+z+t}+\dfrac{z+x}{x+y+z+t}+\dfrac{t+y}{x+y+z+t}=2\)

Do \(1< M< 2\) mà \(1\) và \(2\) là hai số tự nhiên liên tiếp

\(\Rightarrow M\notin\) N

CM: M>1

\(M=\dfrac{x}{x+y+z}+\dfrac{y}{x+y+t}+\dfrac{z}{y+z+t}+\dfrac{t}{x+z+t}\\ >\dfrac{x}{x+y+z+t}+\dfrac{y}{x+y+z+t}+\dfrac{z}{x+y+z+t}+\dfrac{t}{x+y+z+t}=1\left(\text{đ}pcm\right)\)

cm : M<2

\(M< \dfrac{x}{x+y}+\dfrac{y}{x+y}+\dfrac{z}{z+t}+\dfrac{t}{z+t}=1+1=2\left(\text{đ}pcm\right)\)

Vì 1<M<2 nên M không phải là số tự nhiên

Từ \(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}\)

\(\Rightarrow\dfrac{x}{y+z+t}+1=\dfrac{y}{z+t+x}+1=\dfrac{z}{t+x+y}+1=\dfrac{t}{x+y+z}+1\)

\(\Rightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{z+t+x}=\dfrac{x+y+z+t}{t+x+y}=\dfrac{x+y+z+t}{x+y+z}\)

Vì \(x+y+z+t\ne0\) nên ta đi xét \(x+y+z+t=0\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(t+x\right)\\z+t=-\left(x+y\right)\\t+x=-\left(y+z\right)\end{matrix}\right.\). Khi đó

\(P=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=4\)