5x\(=\)8y\(=\)20z và x-y-z\(=\)3
5x\(=\)8y\(=\)20z\(\Rightarrow\) \(\dfrac{5x}{40}=\dfrac{8y}{40}=\dfrac{20z}{40}\Rightarrow\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{8}=\dfrac{y}{5}=\dfrac{z}{2}=\dfrac{x-y-z}{8-5-2}=\dfrac{3}{1}=3\)
\(\left\{{}\begin{matrix}\dfrac{x}{8}=3\Rightarrow x=8.3=24\\\dfrac{y}{5}=3\Rightarrow x=5.3=15\\\dfrac{z}{2}=3\Rightarrow z=2.3=6\end{matrix}\right.\)
Vậy \(x=24;y=15;z=6\)
Từ: \(\dfrac{x}{y}=\dfrac{2}{3}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\)
\(\dfrac{y}{z}=\dfrac{4}{5}\Rightarrow\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\)
\(\rightarrow\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{30}{5}=6\)
\(\left\{{}\begin{matrix}\dfrac{x}{8}=6\Rightarrow x=6.8=48\\\dfrac{y}{12}=6\Rightarrow6.12=72\\\dfrac{z}{15}=6\Rightarrow6.15=90\end{matrix}\right.\)
Vậy \(x=48;y=72;z=90\)
