Đáp án:
Gọi $d=UCLN(3n-2;4n-3)$
$\to\displaystyle\left \{ {{3n-2\qquad\vdots\qquad d} \atop {4n-3\qquad\vdots\qquad d}} \right.$
$\to\displaystyle\left \{ {{4(3n-2)\qquad\vdots\qquad d} \atop {3(4n-3)\qquad \vdots\qquad d}} \right.$
$\to\displaystyle \left \{ {{12n-8\qquad\vdots\qquad d} \atop {12n-9\qquad \vdots\qquad d}} \right.$
$\to 12n-8-(12-9)\qquad\vdots\qquad d$
$\to 1\qquad\vdots \qquad d$
$\to d=1$
$\to \dfrac{3n-2}{4n-3}$ tối giản
