*)Nếu \(x=y=z=t\) \(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{y+x+t}=\dfrac{t}{x+y+z}\) Áp dụng tích chất dãy tỉ số bằng nhau ta có: \(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{y+x+t}=\dfrac{t}{x+y+z}=\dfrac{x+y+z+t}{3\left(x+y+z+t\right)}=\dfrac{1}{3}\)=> \(P=\dfrac{2x}{2x}+\dfrac{2x}{2x}+\dfrac{2x}{2x}+\dfrac{2x}{2x}=4\) *)Nếu có ít nhất 2 số khác nhau , giả sử \(xe y\) => \(\dfrac{x}{y+z+t}=\dfrac{y}{x+z+t}=\dfrac{x-y}{y+z+t-x-z-t}=\dfrac{x-y}{y-x}=-1\) => \(x=-\left(y+z+t\right)\Rightarrow x+y+z+t=0\) => \(\left[{}\begin{matrix}x+y=-\left(z+t\right)\Rightarrow\dfrac{x+y}{z+t}=-1\\y+z=-\left(t+x\right)\Rightarrow\dfrac{y+z}{t+x}=-1\\z+t=-\left(x+y\right)\Rightarrow\dfrac{z+t}{x+y}=-1\\t+x=-\left(y+z\right)\Rightarrow\dfrac{t+x}{y+z}=-1\end{matrix}\right.\) => \(P=-1-1-1-1=-4\) Vậy P=4 P = -4