Ta có: \(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{x+t+y}=\frac{t}{x+y+z}\)
Thêm 1 vào mỗi phân số ta được:
\(\frac{x}{y+z+t}+1=\frac{y}{z+t+x}+1=\frac{z}{x+t+y}+1=\frac{t}{x+y+z}+1\)
\(\Rightarrow\frac{x+y+z+t}{y+z+t}=\frac{x+y+z+t}{z+t+x}=\frac{x+y+z+t}{x+t+y}=\frac{x+y+z+t}{x+y+z}\)
- Nếu x + y + z + t \(\ne\) 0 thì x = y = z = t
\(\Rightarrow P=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}=\frac{x+x}{x+x}+\frac{x+x}{x+x}+\frac{x+x}{x+x}+\frac{x+x}{x+x}=1+1+1+1=4\)
- Nếu x + y + z + t = 0 thì x + y = -(z + t)
y + z = -(t + x)
z + t = -(x + y)
t + x = -(y + z)
\(\Rightarrow P=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}=\frac{-\left(z+t\right)}{z+t}+\frac{-\left(t+x\right)}{t+x}+\frac{-\left(x+y\right)}{x+y}+\frac{-\left(y+z\right)}{y+z}=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
