\(4x=5y=7z\Rightarrow\dfrac{4x}{140}=\dfrac{5y}{140}=\dfrac{7z}{140}=\dfrac{x}{35}=\dfrac{y}{28}=\dfrac{z}{20}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta được :
\(\dfrac{x}{35}=\dfrac{y}{28}=\dfrac{z}{20}=\dfrac{x-y+z}{35-28+20}=\dfrac{54}{27}=2\)
*\(\dfrac{x}{35}=2\Rightarrow x=2.35=70\)
*\(\dfrac{y}{28}=2\Rightarrow y=2.28=56\)
*\(\dfrac{z}{20}=2\Rightarrow z=2.20=40\)
Vậy \(x=70;y=56;z=40\)
\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{y}{7}=\dfrac{z}{5}\)
Nên \(\dfrac{x}{35}=\dfrac{y}{28}=\dfrac{z}{20}\)=\(\dfrac{x-y+z}{35-28+20}\)=\(\dfrac{54}{27}=2\)
suy ra x=70,y=56,z=40
