Question

13. Os números A. B e C foram decompostos num produto de lateres primos
B-2x,5 ×7X11
C-3x5x112
A-2 x 3 x 5
Assim sendo, calcula:
a) m.d.c. (A,B)
d) mm.c. (A, B)
b)m.d.c. (AC)
e) m.m.c. (A.C
c) m.d.c. (A.B.C)
1) m.m.c. (BC)​

Answer 1

a)

$mdc(A,B) =$

$mdc(2\cdot3\cdot5 \;e\; 2\cdot5\cdot7\cdot11) =$

$2\cdot5 = \fcolorbox{black}{green}{10}$

d)

$mmc(A,B) =$

$mmc(2\cdot3\cdot5 \;e\; 2\cdot5\cdot7\cdot11)=</p><p>$

$2 \times 3 \times 5 \times 7 \times 11 = \fcolorbox{black}{green}{2310}$

b)

$mdc(A,C) =$

$mdc(2\cdot3\cdot5 \;e\; 3\cdot5\cdot11²)=</strong></p><p><strong>[tex]mdc(2\cdot3\cdot5 \;e\; 3\cdot5\cdot11²)=$

$3 \times 5 = \fcolorbox{black}{green}{15}$

e)

$mmc(A,C) =$

$mmc(2\cdot3\cdot5 \;e\; 3\cdot5\cdot11²)=</strong></p><p><strong>[tex]mmc(2\cdot3\cdot5 \;e\; 3\cdot5\cdot11²)=$

$2 \times 3 \times 5 \times {11}^{2} = \fcolorbox{black}{green}{3630}$

c)

$mdc(A,B,C) =$

$mdc(2\cdot3\cdot5 \;e\; 2\cdot5\cdot7\cdot11 \;e\; 3\cdot5\cdot11²)=</strong></p><p><strong>[tex]mdc(2\cdot3\cdot5 \;e\; 2\cdot5\cdot7\cdot11 \;e\; 3\cdot5\cdot11²)=$

$= \fcolorbox{black}{green}{5}$

f)

$mmc(B,C) =</strong></p><p><strong>[tex]mmc(B,C) =$

$mmc(2\cdot5\cdot7\cdot11 \;e\; 3\cdot5\cdot11²)=</strong></p><p><strong>[tex]mmc(2\cdot5\cdot7\cdot11 \;e\; 3\cdot5\cdot11²)=$

$2 \times 3 \times 5 \times 7 \times {11}^{2} = \fcolorbox{black}{green}{25410}$

Bons estudos!