Ta có : \(3x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\Rightarrow\dfrac{x}{24}=\dfrac{y}{18}\left(1\right)\)
\(5y=6z\Rightarrow\dfrac{y}{6}=\dfrac{z}{5}\Rightarrow\dfrac{y}{18}=\dfrac{z}{15}\left(2\right)\)
Từ (1);(2) \(\Rightarrow\dfrac{x}{24}=\dfrac{y}{18}=\dfrac{z}{15}\)
Đặt \(\dfrac{x}{24}=\dfrac{y}{18}=\dfrac{z}{15}=k\Rightarrow x=24k;y=18k;z=15k\)
\(\text{Ta có }:x.y.z=24k.18k.15k=30\\ \Rightarrow k^3.6480=30\\ \Rightarrow k^3=\dfrac{1}{216}\\ \Rightarrow k=\dfrac{1}{6}\\ \Rightarrow x=24.\dfrac{1}{6}=4\\ y=18.\dfrac{1}{6}=3\\ z=15.\dfrac{1}{6}=2.5\)
Vậy x = 4 ; y = 3 ; z = 2,5
a) Tìm các số x, y, z biết rằng
3x=4y ; 5y=6z và xyz = 30
Giải
Ta có: 3x = 4y; 5y = 6z \(\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\); \(\dfrac{y}{6}=\dfrac{z}{5}\)
\(\Rightarrow\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}\)
Đặt \(\dfrac{x}{8}=\dfrac{y}{6}=\dfrac{z}{5}=k\) \(\Rightarrow\left\{{}\begin{matrix}x=8k\\y=6k\\z=5k\end{matrix}\right.\)
Có \(xyz=30\) \(\Leftrightarrow\) \(8k.6k.5k=30\)
\(\Rightarrow\) \(240k^3=30\)
\(\Rightarrow\) \(k^3=8\)
\(\Rightarrow\) \(k=\sqrt[3]{8}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=8.2=16\\y=6.2=12\\z=5.2=10\end{matrix}\right.\)
