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

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

+) Nếu \(x+y+z+t\ne0\) thì từ (*) suy ra:
\(y+z+t=z+t+x=t+x+y=x+y+z\)

\(\Rightarrow x=y=z=t\)

\(\Rightarrow P=\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}\) \(\Rightarrow P=1+1+1+1=4\)

+) Nếu \(x+y+z+t=0\) thì \(\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(t+x\right)\\z+t=-\left(x+y\right)\\t+x=-\left(y+z\right)\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{-\left(z+t\right)}{z+t}+\dfrac{-\left(t+x\right)}{t+x}+\dfrac{-\left(x+y\right)}{x+y}+\dfrac{-\left(y+z\right)}{y+z}\)\(\Rightarrow P=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)

Vậy \(P=4\) hoặc \(P=-4\)

dùng tính chất của dãy tỉ số bằng nhau từ đó suy ra x=y=z=t là chứng minh được.
 

mink nghi la 1

ta có \(\frac{x}{x+y+z}>\frac{x}{x+y+z+t}\)

        \(\frac{y}{x+y+t}>\frac{y}{x+y+z+t}\)

       \(\frac{z}{y+z+t}>\frac{z}{x+y+z+t}\)

      \(\frac{t}{x+z+t}>\frac{t}{x+y+z+t}\)

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

ta lại có \(\frac{x}{x+y+z}< \frac{x+t}{x+y+z+t}\)

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

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

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

\(\Rightarrow M>\frac{x+x+y+y+z+z+t+t}{x+y+z+t}=\frac{2x+2y+2z+2t}{x+y+z+t}=2\)

\(\Rightarrow1< M< 2\)

vậy M không phải là số tự nhiên

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

\(CM:\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+m}{b+m};m\in\)N*

Biến đổi tương đương.

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

Vì 1<M<2=> M ko phải stn

đừng chơi với ngu vip

TH1: \(x+y+z+t\ne0\) 

Áp dụng t/c dtsbn ta có:

\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}=\dfrac{x+y+z+t}{3\left(x+y+z+t\right)}=\dfrac{1}{3}\)\(\dfrac{x}{y+z+t}=\dfrac{1}{3}\Rightarrow3x=y+z+t\Rightarrow4x=x+y+z+t\\ \dfrac{y}{z+t+x}=\dfrac{1}{3}\Rightarrow3y=x+z+t\Rightarrow4y=x+y+z+t\\ \dfrac{z}{t+x+y}=\dfrac{1}{3}\Rightarrow3z=x+y+t\Rightarrow4z=x+y+z+t\\ \dfrac{t}{x+y+z}=\dfrac{1}{3}\Rightarrow3t=x+y+z\Rightarrow4t=x+y+z+t\)
\(\Rightarrow4x=4y=4z=4t\\ \Rightarrow x=y=z=t\)

\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\\ =1+1+1+1\\ =4\)

TH2: \(x+y+z+t=0\) 

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

\(P=\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\\ =\dfrac{-\left(z+t\right)}{z+t}+\dfrac{-\left(t+x\right)}{t+x}+\dfrac{-\left(x+y\right)}{x+y}+\dfrac{-\left(y+z\right)}{y+z}\\ =-1-1-1-1\\ =-4\)