a) \(4x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}\Rightarrow\dfrac{x}{35}=\dfrac{y}{28}\)
\(3y=7z\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\Rightarrow\dfrac{y}{28}=\dfrac{z}{12}\)
\(\Rightarrow\dfrac{x}{35}=\dfrac{y}{28}=\dfrac{z}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{35}=\dfrac{y}{28}=\dfrac{z}{12}=\dfrac{x-3y+2z}{35-3.28+2.12}=\dfrac{-25}{-25}=1\)
\(\dfrac{x}{35}=1\Rightarrow x=35\)
\(\dfrac{y}{28}=1\Rightarrow y=28\)
\(\dfrac{z}{12}=1\Rightarrow z=12\)
a: Ta có: 4x=5y
nên \(\dfrac{x}{5}=\dfrac{y}{4}\)
hay \(\dfrac{x}{35}=\dfrac{y}{28}\left(1\right)\)
Ta có: 3y=7z
nên \(\dfrac{y}{7}=\dfrac{z}{3}\)
hay \(\dfrac{y}{28}=\dfrac{z}{12}\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\) suy ra \(\dfrac{x}{35}=\dfrac{y}{28}=\dfrac{z}{12}\)
hay \(\dfrac{x}{35}=\dfrac{3y}{84}=\dfrac{2z}{24}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{35}=\dfrac{3y}{84}=\dfrac{2z}{24}=\dfrac{x-3y+2z}{35-84+24}=\dfrac{-25}{-25}=1\)
Do đó: x=35; y=28; z=12
