\(\dfrac{40}{x-30}=\dfrac{20}{y-15}=\dfrac{28}{z-21}\Leftrightarrow\dfrac{x-30}{40}=\dfrac{y-15}{20}=\dfrac{z-21}{28}\)
\(\Rightarrow\dfrac{x-30}{10}=\dfrac{y-15}{5}=\dfrac{z-21}{7}\)
\(\Rightarrow\dfrac{x}{10}-\dfrac{30}{10}=\dfrac{y}{5}-\dfrac{15}{5}=\dfrac{z}{7}-\dfrac{21}{7}\)
\(\Rightarrow\dfrac{x}{10}-3=\dfrac{y}{5}-3=\dfrac{z}{7}-3\)
\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{5}=\dfrac{z}{7}\)
Đặt: \(\dfrac{x}{10}=\dfrac{y}{5}=\dfrac{z}{7}=t\Rightarrow\left\{{}\begin{matrix}x=10t\\y=5t\\z=7t\end{matrix}\right.\)
\(xyz=22400\Leftrightarrow350t^3=22400\Leftrightarrow t^3=64\Rightarrow t=4\)
\(\Rightarrow\left\{{}\begin{matrix}x=40\\y=20\\z=28\end{matrix}\right.\)
Ta có 40x−30 = 20y−15 = 28z−21 => 40x - 4030= 20y - 2015= 28z- 2821
<=> 40x - 43= 20y - 43 = 28z- 43
<=> 40x = 20y = 28z
Đặt 40x = 20y = 28z= k
Suy ra x = 40k, y = 20k, z = 28k
Khi đó xyz = 40k.20k.28k = 22400k3k3
Theo đề xyz = 22400 suy ra k3k3 = 1 <=> k = ±±1
Với k = 1, ta có x = 40, y = 20, z = 28
Với k = -1, ta có x = -40, y = -20, z = -28
