1)Từ \(\dfrac{2x+y+z+t}{x}=\dfrac{x+2y+z+t}{y}=\dfrac{x+y+2z+t}{z}=\dfrac{x+y+z+2t}{t}\)
\(\Rightarrow\dfrac{2x+y+z+t}{x}-1=\dfrac{x+2y+z+t}{y}-1=\dfrac{x+y+2z+t}{z}-1=\dfrac{x+y+z+2t}{t}-1\)
\(\Rightarrow\dfrac{x+y+z+t}{x}=\dfrac{x+y+z+t}{y}=\dfrac{x+y+z+t}{z}=\dfrac{x+y+z+t}{t}\)
Suy ra \(x+y+z+t=0\) hoặc \(x=y=z=t\)
Bài 2:
Áp dụng tc dãy tỉ số bằng nhau
\(x=\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)