Giải thích các bước giải:
a.Gọi $UCLN(2n+1,2n+3)=d$
$\to\begin{cases}2n+1\quad\vdots\quad d\\2n+3\quad\vdots\quad d\end{cases}$
$\to (2n+3)-(2n+1)\quad\vdots\quad d$
$\to 2\quad\vdots\quad d$
$\to d=1$ vì $2n+1$ lẻ
$\to (2n+1,2n+3)=1$
$\to A=\dfrac{2n+1}{2n+3} $ tối giản
b.Gọi UCLN(n+1,3n+4)=d
$\to\begin{cases}n+1\quad\vdots\quad d\\3n+4\quad\vdots\quad d\end{cases}$
$\to 3n+4-3(n+1)\quad\vdots\quad d$
$\to 3n+4-(3n+3)\quad\vdots\quad d$
$\to 1\quad\vdots\quad d$
$\to d=1$
$\to (n+1,3n+4)=1$
$\to B=\dfrac{n+1}{3n+4}$ tối giản
c.Gọi UCLN(2n+3,3n+5)=d
$\to\begin{cases}2n+3\quad\vdots\quad d\\3n+5\quad\vdots\quad d\end{cases}$
$\to 2(3n+5)-3(2n+3)\quad\vdots\quad d$
$\to (6n+10)-(6n+9)\quad\vdots\quad d$
$\to 1\quad\vdots\quad d$
$\to d=1$
$\to (2n+3,3n+4)=1$
$\to C=\dfrac{2n+3}{3n+5}$ tối giản
