\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}\)
\(\dfrac{y}{5}=\dfrac{z}{7}\Rightarrow\dfrac{y}{20}=\dfrac{z}{28}\)
\(\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}=\dfrac{2x}{30}+\dfrac{3y}{60}+\dfrac{-z}{-28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{-124}{62}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=-2\\\dfrac{y}{20}=-2\\\dfrac{z}{28}=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=15.\left(-2\right)=-30\\y=20.\left(-2\right)=-40\\z=28.\left(-2\right)=-56\end{matrix}\right.\)
Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{x}{15}=\dfrac{y}{20}\left(1\right)\)
\(\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{y}{20}=\dfrac{z}{28}\left(2\right)\)
Từ (1) và (2) suy ra:
\(\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{28}\Rightarrow\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{2x}{30}=\dfrac{3y}{60}=\dfrac{z}{28}=\dfrac{2x+3y-z}{30+60-28}=\dfrac{-124}{62}=-2\)
Do đó:
\(\dfrac{x}{15}=-2\Rightarrow x=15.\left(-2\right)=-30\)
\(\dfrac{y}{20}=-2\Rightarrow y=20.\left(-2\right)=-40\)
\(\dfrac{z}{28}=-2\Rightarrow z=28.\left(-2\right)=-56\)
Vậy x = -30; y = -40; z = -56.
\(#Tmiamm\)
