Đáp án:
Giải thích các bước giải:
$a)$ $\frac{n+1}{2n+3}$
Gọi $UC(n+1;2n+3)=d$
$⇒n+1$ $\vdots$ $d$ $2n+3$ $\vdots$ $d$
$⇒2n+2$ $\vdots$ $d$
$⇒(2n+3)-(2n+2)$ $\vdots$ $d$
$⇒1$ $\vdots$ $d$
$⇒d=${$1;-1$}
Vậy $\frac{n+1}{2n+3}$ là phân số tối giản
$b)$ $\frac{2n+3}{4n+8}$
Gọi $UC(2n+3;4n+8)=d$
$⇒2n+3$ $\vdots$ $d$; $4n+8$ $\vdots$ $d$
$⇒4n+6$ $\vdots$ $d$
$⇒(4n+8)-(4n+6)$ $\vdots$ $d$
$⇒2$ $\vdots$ $d$
$⇒d=${$2;1;-1;-2$}
Mà $2n+3$ $\not\vdots$ 2
$⇒d=${$1;-1$}
Vậy ...
$c)$ $\frac{3n+2}{5n+3}$
Gọi $UC(3n+2;5n+3)=d$
$⇒3n+2$ $\vdots$ $d$; $5n+3$ $\vdots$ $d$
$⇒15n+10$ $\vdots$ $d$; $15n+9$ $\vdots$ $d$
$⇒(15n+10)-(15n+9)$ $\vdots$ $d$
$⇒1$ $\vdots$ $d$
$⇒d=${$1;-1$}
Vậy...
