Gọi 2 số dương đó lần lượt là a và b
Ta có: \(20\left(a+b\right)=140\left(a-b\right)=7ab\)
\(\Leftrightarrow\dfrac{20\left(a+b\right)}{140}=\dfrac{140\left(a-b\right)}{140}=\dfrac{7ab}{140}\)
\(\Leftrightarrow\dfrac{a+b}{7}=\dfrac{a-b}{1}=\dfrac{ab}{20}\)
Xét: \(\dfrac{a+b}{7}=\dfrac{a-b}{1}=\dfrac{a+b+a-b}{7+1}=\dfrac{a+b-a+b}{7-1}=\dfrac{2a}{8}=\dfrac{2b}{6}=\dfrac{a}{4}=\dfrac{b}{3}\)
Hay: \(\dfrac{a}{4}=\dfrac{b}{3}=\dfrac{ab}{20}\Leftrightarrow\left\{{}\begin{matrix}20a=4ab\\20b=3ab\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=5\\a=\dfrac{20}{3}\end{matrix}\right.\)