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\(27^{13}=\left(3^3\right)^{13}=3^{39};243^8=\left(3^5\right)^8=3^{40};3^{39}< 3^{40}\Rightarrow27^{13}< 243^8\\ 125^{80}=\left(5^3\right)^{80}=5^{240};25^{118}=\left(5^2\right)^{118}=5^{236};5^{240}>5^{236}\Rightarrow125^{80}>25^{118}\)
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
\(a,ƯC\left(40,24\right)=Ư\left(8\right)=\left\{...\right\}\\ b,ƯC\left(12,52\right)=Ư\left(4\right)=\left\{...\right\}\\ c,ƯC\left(36,990\right)=Ư\left(18\right)=\left\{...\right\}\\ d,ƯC\left(54,36\right)=Ư\left(9\right)=\left\{...\right\}\\ e,ƯC\left(10,20,70\right)=Ư\left(10\right)=\left\{...\right\}\\ f,ƯC\left(25,55,75\right)=Ư\left(5\right)=\left\{...\right\}\\ g,ƯC\left(80,144\right)=Ư\left(16\right)=\left\{...\right\}\\ h,ƯC\left(63,2970\right)=Ư\left(9\right)=\left\{...\right\}\\ i,ƯC\left(65,125\right)=Ư\left(5\right)=\left\{...\right\}\\ j,ƯC\left(9,18,72\right)=Ư\left(9\right)=\left\{...\right\}\\ k,ƯC\left(24,36,60\right)=Ư\left(12\right)=\left\{...\right\}\\ l,ƯC\left(16,42,86\right)=Ư\left(2\right)=\left\{..\right\}\)
a: 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: \(=11+\dfrac{3}{13}-2-\dfrac{4}{7}-5-\dfrac{3}{13}=4-\dfrac{4}{7}=\dfrac{24}{7}\)
b: \(=\dfrac{11}{2}\cdot\dfrac{15}{4}=\dfrac{165}{8}\)
c: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}-6-\dfrac{2}{9}=6+\dfrac{3}{5}=\dfrac{33}{5}\)
d: \(=6+\dfrac{4}{9}+3+\dfrac{7}{11}-4-\dfrac{4}{9}=5+\dfrac{7}{11}=\dfrac{62}{11}\)
\(125^{80}=\left(5^3\right)^{80}=5^{240}\)
\(25^{48}=\left(5^2\right)^{48}=5^{96}\)
Vì 240 > 96 mà 5 = 5
=> \(5^{240}>5^{96}\)
=> \(125^{80}>25^{48}\)
12580 và 2548
12580=( 53) 80 =5240
2548 = (52)48 =596
=>5240 >596=>12580 > 2548