Đáp án:
Giải thích các bước giải:
a, $\frac{2n+1}{3n+1}$
Gọi ƯCLN( 2n+1 , 3n+1 ) = d
$⇒\left \{ {{2n+1 \vdots d} \atop {3n+1\vdots d}} \right.$
$⇒\left \{ {{3(2n+1) \vdots d} \atop {2(3n+1)\vdots d}} \right.$
$⇒\left \{ {{6n+3 \vdots d} \atop {6n+2\vdots d}} \right.$
$⇒ ( 6n + 3 ) - ( 6n + 2 ) \vdots d $
$⇒ 1 \vdots d $
$⇒ d = 1 $
$⇒ ƯCLN( 2n+1, 3n+1 ) = 1 $
$⇒ \frac{2n+1}{3n+1}$ tối giản
b, $\frac{2n+1}{4n+3}$
gọi ƯCLN( 2n + 1 . 4n + 3 ) = d
$⇒\left \{ {{2n+1 \vdots d} \atop {4n+3\vdots d}} \right.$
$⇒\left \{ {{2(2n+1) \vdots d} \atop {4n+3\vdots d}} \right.$
$⇒\left \{ {{4n+2 \vdots d} \atop {4n+3\vdots d}} \right.$
$⇒ ( 4n+3 ) - ( 4n + 2) \vdots d $
$⇒ 1 \vdots d $
$⇒ d = 1 $
$⇒ ƯCLN( 2n+1 , 4n+3 ) = 1 $
$⇒ \frac{2n+1}{4n+3} $ tối giản
c, $\frac{4n+2}{12n+7}$
Gọi $ƯCLN( 4n+2 , 12n+7 ) = d $
$⇒\left \{ {{4n+2 \vdots d} \atop {12n+7\vdots d}} \right.$
$⇒\left \{ {{3(4n+2) \vdots d} \atop {12n+7\vdots d}} \right.$
$⇒\left \{ {{12n+6 \vdots d} \atop {12n+7\vdots d}} \right.$
$⇒ ( 12n + 7 ) - ( 12n + 6 ) \vdots d $
$⇒ 1 \vdots d $
$⇒ d = 1 $
$⇒ ƯCLN( 4n+2 , 12n+7 ) = 1 $
$⇒ \frac{4n+2}{12n+7}$ tối giản
