\(1\dfrac{3}{7}=\dfrac{10}{7}\)
Theo đề ta có :
\(a:\dfrac{3}{5}\in N\Leftrightarrow a.\dfrac{5}{3}\in N\left(1\right)\)
\(a:\dfrac{10}{7}\in N\Leftrightarrow a.\dfrac{7}{10}\in N\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\)\(\left\{{}\begin{matrix}5a⋮3\\7a⋮10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a⋮3\\a⋮10\end{matrix}\right.\) (do \(ƯCLN\left(5,3\right)=1;ƯCLN\left(7,10\right)=1\))
Mà a nhỏ nhất
\(\Leftrightarrow a\in BCNN\left(3,10\right)=30\)
Giải
\(BCNN_{\left(\dfrac{3}{5};1\dfrac{3}{7}\right)}=\dfrac{6}{7}\)
