đề là gì vậy bạn

Gọi 199010+19909 là A

Gọi 199110 là B

A=199010+19909=19909(1990+1)=19909.1991

B=199110=19919.1991

Vậy A<B

ta có

199010+19909=218929

199110=199110

Suy ra 199010+19909>199110

a: \(2^{300}=8^{100}\)

\(3^{200}=9^{100}\)

mà 8<9

nên \(2^{300}< 3^{200}\)

b: \(3^{500}=243^{100}\)

\(7^{300}=343^{100}\)

mà 243<243

nên \(3^{500}< 7^{300}\)

Đức Huy trả lời sai vì XXIVI gồm XXI và VI = 21 + 6 = 27

A) IXMIXCIX

B) 1 :27

    2: 25

C) IXXMIXMIXCIXXIX

C, so la ma lon nhat co ba chu so la xxxxx do !!!

Dễ nhưng dài e nên tách ra

a) ƯCLN(12;18) = 6

b) ƯCLN(12;10)=2

c) ƯCLN(24;48)=24

d) ƯCLN(300;280)=20

e) ƯCLN(9;81)=9

f) ƯCLN(11;15)=1

g) ƯCLN(1;10)=1

h) ƯCLN(150;84)=6

i) ƯCLN(46;138)=46

j) ƯCLN(32;192)=32

k) ƯCLN(18;42)=6

l) ƯCLN(28;48)=4

m) ƯCLN(24;36;60) = 12

n) ƯCLN(12;15;10)=1

p) ƯCLN(16;32;112) = 16

q) ƯCLN(14;82;124) = 2

r) ƯCLN(25;55;75) = 5

a: \(42=2\cdot3\cdot7;70=2\cdot5\cdot7\)

=>\(BCNN\left(42;70\right)=2\cdot3\cdot5\cdot7=210\)

=>\(BC\left(42;70\right)=B\left(210\right)=\left\{0;210;420;...\right\}\)

b: \(70=2\cdot5\cdot7;180=3^2\cdot5\cdot2^2\)

=>\(BCNN\left(70;180\right)=2^2\cdot3^2\cdot5\cdot7=1260\)

=>\(BC\left(70;180\right)=\left\{1260;2520;...\right\}\)

c: \(5=5;7=7;8=2^3\)

=>\(BCNN\left(5;7;8\right)=5\cdot7\cdot8=280\)

=>\(BC\left(5;7;8\right)=\left\{280;560;...\right\}\)

d: \(12=2^2\cdot3;18=3^2\cdot2\)

=>\(BCNN\left(12;18\right)=2^2\cdot3^2=36\)

=>\(BC\left(12;18\right)=\left\{36;72;...\right\}\)

e: \(15=3\cdot5;18=3^2\cdot2\)

=>\(BCNN\left(15;18\right)=3^2\cdot2\cdot5=90\)

=>\(BC\left(15;18\right)=\left\{90;180;...\right\}\)

f: \(84=2^2\cdot3\cdot7;108=3^3\cdot2^2\)

=>\(BCNN\left(84;108\right)=2^2\cdot3^3\cdot7=756\)

=>\(BC\left(84;108\right)=\left\{756;1512;...\right\}\)

j: \(33=3\cdot11;44=2^2\cdot11;55=5\cdot11\)

=>\(BCNN\left(33;44;55\right)=3\cdot2^2\cdot5\cdot11=660\)

=>\(BC\left(33;44;55\right)=\left\{660;1320;...\right\}\)

g: \(1=1;12=2^2\cdot3;27=3^3\)

=>\(BCNN\left(1;12;27\right)=1\cdot2^2\cdot3^3=108\)

=>\(BC\left(1;12;27\right)=\left\{108;216;...\right\}\)

n: \(5=5;9=3^2;11=11\)

=>\(BCNN\left(5;9;11\right)=5\cdot3^2\cdot11=495\)

=>\(BC\left(5;9;11\right)=\left\{495;990;...\right\}\)

24 = 2³.3;  36 = 2⁴.3⁴; 60 = 2².3.5

ƯCLN( 24; 36; 60) = 2².3 = 12

12 = 2².3; 15 = 3.5; 10 = 2.5 

ƯCLN(12; 15; 10) = 1

24  = 2³.3; 16 = 2⁴; 8 = 2³

ƯCLN(24; 16; 8) = 2³

9 = 3²; 81 = 3⁴ 

ƯCLN( 9; 81) =  9

11 = 11; 15 = 3.5

ƯCLN( 11; 15) = 1

1 = 1; 10 = 2.5

ƯCLN(1; 10) = 1

150 = 2.3.5²;  84 = 2².3.7 

ƯCLN( 150; 84) = 6

46 = 2.23; 138 = 2.3.23 

ƯCLN(46; 138) = 2.23 = 46

16 = 2⁴; 32 = 2⁵; 124 = 2².31

ƯCLN( 16; 32; 124) = 2² = 4

14 = 2.7; 82 = 2.41; 124 = 2².31 

ƯCLN( 14; 82; 124) = 2

a, 5 và 8 =40 ; 12 và 18 = 36 ; 7 và 14 =14 ; 12 và 19 =228

b, 15 , 25 = 75 ; 30,60 = 60 ; 75 và 15 =75 ; 25,5 =25

\(a,ƯC\left(40,24\right)=Ư\left(8\right)=\left\{...\right\}\\ b,ƯC\left(12,52\right)=Ư\left(4\right)=\left\{...\right\}\\ c,ƯC\left(36,990\right)=Ư\left(18\right)=\left\{...\right\}\\ d,ƯC\left(54,36\right)=Ư\left(9\right)=\left\{...\right\}\\ e,ƯC\left(10,20,70\right)=Ư\left(10\right)=\left\{...\right\}\\ f,ƯC\left(25,55,75\right)=Ư\left(5\right)=\left\{...\right\}\\ g,ƯC\left(80,144\right)=Ư\left(16\right)=\left\{...\right\}\\ h,ƯC\left(63,2970\right)=Ư\left(9\right)=\left\{...\right\}\\ i,ƯC\left(65,125\right)=Ư\left(5\right)=\left\{...\right\}\\ j,ƯC\left(9,18,72\right)=Ư\left(9\right)=\left\{...\right\}\\ k,ƯC\left(24,36,60\right)=Ư\left(12\right)=\left\{...\right\}\\ l,ƯC\left(16,42,86\right)=Ư\left(2\right)=\left\{..\right\}\)

a:UC(40;24)=Ư(8)

b: UC(12;52)=Ư(4)

tui cần cả các bước giải nữa

a: 18=3^2*2; 42=2*3*7

=>ƯCLN(18;42)=3*2=6

b: 28=2^2*7

48=2^4*3

=>ƯCLN(28;48)=2^2=4

c: 24=2^3*3

36=2^2*3^2

60=2^2*3*5

=>ƯCLN(24;36;60)=12

d: 12=2^2*3

15=3*5

10=2*5

=>ƯCLN(12;15;10)=1

e: 24=2^3*3

16=2^4

8=2^3

=>ƯCLN(24;16;8)=2^3=8

h: 25=5^2; 55=5*11; 75=5^2*3

=>ƯCLN(25;55;75)=5

\(a,4\dfrac{1}{4}-2=\dfrac{17}{4}-2=\dfrac{9}{4};\dfrac{5}{8}+2\dfrac{3}{5}=\dfrac{5}{8}+\dfrac{13}{5}=\dfrac{129}{40}\\ b,4\dfrac{4}{9}:2=\dfrac{40}{9}\cdot\dfrac{1}{2}=\dfrac{20}{9};\dfrac{2}{3}+3\dfrac{1}{6}=\dfrac{2}{3}+\dfrac{19}{6}=\dfrac{23}{6}\\ c,3\dfrac{1}{5}+2=\dfrac{9}{5}+2=\dfrac{19}{5};\dfrac{3}{5}-2\dfrac{4}{5}=\dfrac{3}{5}-\dfrac{14}{5}=-\dfrac{11}{5}\)

\(d,5\dfrac{1}{7}-2=\dfrac{36}{7}-2=\dfrac{22}{7};\dfrac{4}{5}:1\dfrac{1}{5}=\dfrac{4}{5}:\dfrac{6}{5}=\dfrac{2}{3}\\ e,2\dfrac{3}{5}+1=\dfrac{13}{5}+1=\dfrac{18}{5};\dfrac{1}{4}\cdot2\dfrac{2}{3}=\dfrac{1}{4}\cdot\dfrac{8}{3}=\dfrac{2}{3}\\ f,4\dfrac{1}{3}\cdot1=\dfrac{13}{3};\dfrac{1}{2}+5\dfrac{2}{7}=\dfrac{1}{2}+\dfrac{37}{7}=\dfrac{81}{14}\)