Giải thích các bước giải:
Bài 4:
a.Gọi $UCLN(15n+1,30n+1)=d$
$\to\begin{cases}15n+1\quad\vdots\quad d\\30n+1\quad\vdots\quad d\end{cases}$
$\to 2(15n+1)-(30n+1)\quad\vdots\quad d$
$\to 1\quad\vdots\quad d$
$\to d=1$
$\to (15n+1,30n+1)=1$
$\to \dfrac{15n+1}{30n+1}$ tối giản
b.Gọi $UCLN(n^3+2n,n^4+3n^2+1)=d$
$\to\begin{cases}n^3+2n\quad\vdots\quad d\\ n^4+3n^2+1\quad\vdots\quad d\end{cases}$
$\to (n^4+3n^2+2)-n(n^3+2n)\quad\vdots\quad d$
$\to n^2+2\quad\vdots\quad d$
$\to n^4+3n^2+1=(n^4-2^2)+3(n^2+2)-1\quad\vdots\quad d$
$\to n^4+3n^2+1=(n^2-2)(n^2+2)+3(n^2+2)-1\quad\vdots\quad d$
$\to 1\quad\vdots\quad d$
$\to d=1$
$\to UCLN(n^3+2n,n^4+3n^2+1)=1$
$\to \dfrac{n^3+2n}{n^4+3n^2+1}$ tối giản
