Ta có: \(3x=y\Rightarrow x=\dfrac{y}{3}\Rightarrow\dfrac{x}{4}=\dfrac{y}{12}\)
\(5y=4z\Rightarrow\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\)
\(\Rightarrow\dfrac{x}{4}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{6x}{24}=\dfrac{7y}{84}=\dfrac{8z}{120}=\dfrac{6x+7y+8z}{24+84+120}=\dfrac{456}{228}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.4=8\\y=2.12=24\\z=2.15=30\end{matrix}\right.\)
Ta có: 3x=y
nên \(\dfrac{x}{1}=\dfrac{y}{3}\)
hay \(\dfrac{x}{4}=\dfrac{y}{12}\left(1\right)\)
Ta có: 5y=4z
nên \(\dfrac{y}{4}=\dfrac{z}{5}\)
hay \(\dfrac{y}{12}=\dfrac{z}{15}\left(2\right)\)
Từ (1) và (2) suy ra \(\dfrac{x}{4}=\dfrac{y}{12}=\dfrac{z}{15}\)
hay \(\dfrac{6x}{24}=\dfrac{7y}{84}=\dfrac{8z}{120}\)
mà 6x+7y+8z=456
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{6x}{24}=\dfrac{7y}{84}=\dfrac{8z}{120}=\dfrac{6x+7y+8z}{24+84+120}=\dfrac{456}{228}=2\)
Do đó: x=8; y=24; z=30