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 \(P=-4\)
