\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\left(\dfrac{x}{2}\right)^3=\dfrac{xyz}{2.3.5}\Leftrightarrow\dfrac{x^3}{8}=\dfrac{240}{30}=8\Leftrightarrow x^3=64\Leftrightarrow x=4\Rightarrow\left\{{}\begin{matrix}y=6\\z=10\end{matrix}\right.\)
Đặt\(\dfrac{x}{2}\)=\(\dfrac{y}{3}\)=\(\dfrac{z}{5}\)=k
=>x = 2k; y = 3k; z =5k
Mà x.y.z=240 => 2k.3k.5k=240
=>(k.k.k).(2.3.5)=240
=> \(k^3\) . 30 =240
=> \(k^3\) =240: 30
=> \(k^3\) = 8
=> k = \(\pm\) 2
Từ k=2 => x=2.2=4
k=-2=> x=-2.2=-4
Từ k=2 => y=2.3
k=-2=> y=-2.3=-6
Từ k=2=> z=2.5=10
k=-2=> z=-2.5=-10
Vậy x\(\in\pm\) 4
y\(\in\pm\) 6
z\(\in\pm\) 10
tìm x,y,z biết: x/2= y/3= z/5 với x.y.z=240
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)
=> x = 2k, y = 3k, z = 5k
Ta có : x . y . z = 2k . 3k . 5k = 240
=> x . y . z = \(30k^3\)= 240
=> \(k^3=230:40=8\)
=> k = 2
=> x = 2k = 2 . 2 =4
=> y = 3k = 3 . 2 = 6
=> z = 5 . 2 = 10
Vậy x = 4, y = 6, z = 10