$\begin{array}{l}a)\,\,\text{- Gọi $Ư CLN(n+3,n+4)=d\,\,(d \in \mathbb{Z})$}\\\to \begin{cases} n+3\,\,\vdots\,\,d\\n+4\,\,\vdots\,\,d\end{cases}\\\to (n+3)-(n+4)\,\,\vdots\,\,d\\\to(n-n)+(3-4)\,\,\vdots\,\,d\\\to-1\,\,\vdots\,\,d\\\to d=\pm1\\\to\text{$n+3$ và $n+4$ nguyên tố cùng nhau}\\\to \text{$\dfrac{n+3}{n+4}$ tối giản}\\\,\\b)\,\,\text{- Gọi $ƯC LN(3n-2,4n-3)=d\,\,(d \in \mathbb{Z})$}\\\to \begin{cases} 3n-2\,\,\vdots\,\,d\\4n-3\,\,\vdots\,\,d\end{cases}\\\to \begin{cases} 4(3n-2)\,\,\vdots\,\,d\\3(4n-3)\,\,\vdots\,\,d\end{cases}\\\to\begin{cases} 12n-8\,\,\vdots\,\,d\\12n-9\,\,\vdots\,\,d\end{cases}\\\to (12n-8)-(12n-9)\,\,\vdots\,\,d\\\to (12n-12n)+(-8+9)\,\,\vdots\,\,d\\\to1\,\,\vdots\,\,d\\\to d=\pm1\\\to\text{$3n-2$ và $4n-3$ nguyên tố cùng nhau}\\\to \dfrac{3n-2}{4n-3}\text{ tối giản}\\\,\\c)\,\,\text{- Gọi $ƯC LN(2n+5,3n+7)=d\,\,(d \in \mathbb{Z})$}\\\to \begin{cases} 2n+5\,\,\vdots\,\,d\\3n+7\,\,\vdots\,\,d\end{cases}\\\to \begin{cases} 3(2n+5)\,\,\vdots\,\,d\\2(3n+7)\,\,\vdots\,\,d\end{cases}\\\to\begin{cases} 6n+15\,\,\vdots\,\,d\\6n+14\,\,\vdots\,\,d\end{cases}\\\to (6n+15)-(6n+14)\,\,\vdots\,\,d\\\to (6n-6n)+(15-14)\,\,\vdots\,\,d\\\to1\,\,\vdots\,\,d\\\to d=\pm1\\\to\text{$2n+5$ và $3n+7$ nguyên tố cùng nhau}\\\to \dfrac{2n+5}{3n+7}\text{ tối giản} \end{array}$
