Sửa đề: \(\frac{2x+y+z+t}{x}=\frac{x+2y+z+t}{y}=\frac{x+y+2z+t}{z}=\frac{x+y+z+2t}{t}\)
Ta có: \(\frac{2x+y+z+t}{x}=\frac{x+2y+z+t}{y}=\frac{x+y+2z+t}{z}=\frac{x+y+z+2t}{t}\)
=>\(\frac{2x+y+z+t}{x}-2=\frac{x+2y+z+t}{y}-2=\frac{x+y+2z+t}{z}-2=\frac{x+y+z+2t}{t}-2\)
=>\(\frac{y+z+t}{x}=\frac{x+z+t}{y}=\frac{x+y+t}{z}=\frac{x+y+z}{t}\)
=>\(\frac{y+z+t}{x}+1=\frac{x+z+t}{y}+1=\frac{x+y+t}{z}+1=\frac{x+y+z}{t}+1\)
=>\(\frac{x+y+z+t}{x}=\frac{x+y+z+t}{y}=\frac{x+y+z+t}{z}=\frac{x+y+z+t}{t}\) (1)
TH1: x+y+z+t=0
=>x+y=-(z+t); y+z=-(t+x)
\(A=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}\)
\(=\frac{-\left(z+t\right)}{z+t}+\frac{-\left(x+t\right)}{x+t}+\frac{-\left(x+y\right)}{x+y}+\frac{-\left(y+z\right)}{y+z}\)
=-1-1-1-1
=-4
TH2: x+y+z+t<>0
(1) sẽ tương ứng với \(\frac{1}{x}=\frac{1}{y}=\frac{1}{z}=\frac{1}{t}\)
=>x=y=z=t
\(A=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}\)
\(=\frac{x+x}{x+x}+\frac{x+x}{x+x}+\frac{x+x}{x+x}+\frac{x+x}{x+x}\)
=1+1+1+1
=4
