Gọi d là \(UCLN\left(25m+7;15m+4\right)\)
\(\Rightarrow\left\{{}\begin{matrix}25m+7⋮d\\15m+4⋮d\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3\left(25m+7\right)⋮d\\5\left(15m+4\right)⋮d\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}75m+21⋮d\\75m+20⋮d\end{matrix}\right.\)
\(\Rightarrow\left[\left(75m+21\right)-\left(75m+20\right)\right]⋮d\)
\(\Rightarrow1⋮d\Rightarrow d=1\)
Vậy \(\dfrac{25m+7}{15m+4}\) tối giản \(\forall m\in Z\)