a) 28 = 2 2 . 7 ; 70 = 2.5.7 .Vậy ƯCLN(28,70) = 2.7 = 14

b) ƯCLN(36, 60,72) =12

c) ƯCLN(60,180) = 60

d) ƯCLN(28, 39, 35) = 1

a) 40 = 2³.5

60 = 2².3.5

ƯCLN(40; 60) = 2².5 = 20

ƯC(40; 60) = Ư(20) = {1; 2 ; 4; 5; 10; 20}

b) 28 = 2².7

39 = 3.13

35 = 5.7

ƯCLN(28; 39; 35) = 1

ƯC(28; 39; 35) = Ư(1) = 1

c) 48 = 2⁴.3

60 = 2².3.5

120 = 2³.3.5

ƯCLN(48; 60; 120) = 2².3 = 12

ƯC(48; 60; 120) = Ư(12) = {1; 2; 3; 4; 6; 12}

d) 30 = 2.3.5

75 = 3.5²

135 = 3³.5

ƯCLN(30; 75; 135) = 3.5 = 15

ƯC(30; 75; 135) = Ư(15) = {1; 3; 5; 15}

a) 20

b) ko pk là 39 mà là 49 ⇒ ƯCLN là 7

c)12

d)15

a) Ta có: 17 = 17 và 25 = 5 2 . Vậy ƯCLN(17,25) = 1

b) Ta có: 135 ⋮ 15. Vậy ƯCLN(15,135) = 15

c) Ta có: 96 ⋮ 24 và 264 ⋮ 24. Vậy ƯCLN(24,96,264) = 24

d) Ta có: 28 = 2 2 . 7 ; 39 = 3.13 và 35 = 5.7. Vậy ƯCLN(45,105) = 3.5 = 15

a) x – 12 = 14

x = 14 +12

x = 26

b) 2x – 13 = 3.17

2x = 51 + 13

2x = 64

x = 64 : 2

x = 32

c) x – 43 = 2.18

x = 36 +43

x = 79

d) (x – 14).39 = 0

x – 14 = 0

x = 14

e) (13 – x).28 = 28

13 – x = 1

13 – x = 1

x = 13 – 1

x = 12

f) 22.(35 – x) = 22

35 – x = 1

x = 35 – 1

x = 34

g) x – 24 : 2 = 18

x – 12 = 18

x = 39

h) 400 + (275 – x) = 570

275 – x = 570 – 400

275 – x = 170

x = 275 – 170

x = 105

a) 40=2³.5

60=2².3.5; ƯCLN(40,60)=2².5=20

b)36=2².3²

60=2².3.5; ƯCLN(36,60)=2².3=12

c)13=13

20=2².5; ƯCLN(13,20)=1

d)28=2².7

39=3.13

35=5.7; ƯCLN(28,39,35)=1

a) 40=2³.5

60=2².3.5; ƯCLN(40,60)=2².5=20

b)36=2².3²

60=2².3.5; ƯCLN(36,60)=2².3=12

c)13=13

20=2².5; ƯCLN(13,20)=1

d)28=2².7

a: \(-\frac{10}{3}+\frac{13}{10}-\frac16+\frac{7}{10}\)

\(=\left(-\frac{10}{3}-\frac16\right)+\left(\frac{13}{10}+\frac{7}{10}\right)\)

\(=-\frac{21}{6}+\frac{20}{10}=-\frac72+2=-\frac32\)

b: \(1-\frac{20}{6}+\frac{39}{30}-\frac{4}{24}+\frac{34}{50}\)

\(=1-\frac{20}{6}+\frac16+1,3+0,68\)

\(=1-\frac{19}{6}+1,98=2,98-\frac{19}{6}=\frac{298}{100}-\frac{19}{6}=\frac{-56}{300}=\frac{-14}{75}\)

c: \(\frac{15}{24}-\frac{6}{26}+\frac{28}{48}-\frac{30}{39}+\frac{10}{24}\)

\(=\left(\frac{15}{24}+\frac{10}{24}+\frac{28}{48}\right)-\left(\frac{3}{13}+\frac{10}{13}\right)\)

\(=\left(\frac{25}{24}+\frac{14}{24}\right)-1=\frac{39}{24}-1=\frac{15}{24}=\frac58\)

d: \(-\frac{5}{10}+\frac{12}{28}-\frac{3}{27}-\frac{7}{18}+\frac{20}{35}\)

\(=-\frac12+\frac37-\frac19-\frac{7}{18}+\frac47\)

\(=\left(-\frac12-\frac19-\frac{7}{18}\right)+\left(\frac37+\frac47\right)=-1+1=0\)