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

\(Tương\)\(tự\):\(\Rightarrow M< \dfrac{2\left(x+y+z+t\right)}{x+y+z+t}\)

\(Ta\) \(có\):\(2>M>1\)

\(\Rightarrow M\notin N\)\(sao\)

Ta có:

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

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

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

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

Cộng vế với vế ta được:

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

\(\Rightarrow1< \dfrac{x}{x+y+z}+\dfrac{y}{x+y+t}+\dfrac{z}{y+z+t}+\dfrac{t}{x+z+t}< 2\)

\(\Rightarrow1< M< 2\)

=> M không là số tự nhiên

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

Giả sử \(A\in N\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+y+z}\in N\\\dfrac{y}{x+y+t}\in N\\\dfrac{z}{y+z+t}\in N\\\dfrac{t}{x+z+t}\in N\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x⋮x+y+z\\y⋮x+y+t\\z⋮y+z+t\\t⋮x+z+t\end{matrix}\right.\)

Vì \(x;y;z;t\in N\circledast\) nên:

\(\left\{{}\begin{matrix}x\ge x+y+z\\y\ge x+y+t\\z\ge y+z+t\\t\ge x+z+t\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-x\ge x+y+z-x\\y-y\ge x+y+t-y\\z-z\ge y+z+t-z\\t-t\ge x+z+t-t\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+z\le0\\x+t\le0\\y+t\le0\\x+z\le0\end{matrix}\right.\)

Vì \(x;y;z;t\in N\circledast\) nên những điều trên không thể xảy ra

\(\Rightarrow\) điều giả sử sai,\(A\notin N\left(đpcm\right)\)

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

\(\dfrac{a}{b}< 1\Rightarrow\) \(\dfrac{a}{b}< \dfrac{a+m}{b+m}\) (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