Gọi `ƯCLN_((3n+4; n+1))=d`
$\begin{array}{l}\Rightarrow\left\{\begin{matrix}3n+4 \vdots d& \\n+1 \vdots d& \end{matrix}\right.\Rightarrow\left\{\begin{matrix}3n+4 \vdots d& \\3(n+1) \vdots d& \end{matrix}\right. \Rightarrow \left\{\begin{matrix}3n+4 \vdots d& \\3n+3 \vdots d& \end{matrix}\right.\end{array}$
`=>(3n+4)-(3n+3)\vdots d`
`=>3n+4-3n-3\vdots d`
`=>1\vdots d`
`=>d=1`
`=>ƯCLN_((3n+4; n+1))=1`
Vậy phân số `(3n+4)/(n+1)` tối giản