199^20 < 2003^15

199²⁰<2003¹⁵

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a) Ta có:

\(199^{20}=\left[\left(199\right)^4\right]^5=1568239201^5\)

\(2003^{15}=\left[\left(2003\right)^3\right]^5=8036054027^5\)

Mà: \(8036054027>1568239201\)

\(\Rightarrow1568239201^5< 8036054027^5\) 

\(\Rightarrow199^{20}< 2003^{15}\)

b) Xem lại đề 

a, $5^{3} =5\times5\times5=125$

$3^{5} =3\times3\times3=27$

$125>27=>5^{3}>3^{5}$

$3^{2}=3\times3=9$

$2^{3}=2\times2\times2=8$

$9>8=>3^{2}>2^{3}$

$2^{6} =2\times2\times2\times2\times2\times2=64$

$6^{2}=6\times6=36$

$64>36=>2^{6}>6^{2}$

b, $2015\times2017=2015\times(2016+1)=2015\times2016+2015$

$2016^{2}=2016\times2016=2016\times(2015+1)=2016\times2015+2016$

$2015\times2016+2015<2016\times2015+2016=>2015\times2017<2016^{2}$

c, $199^{20}=199^{4\times5}=(199^{4})^{5}= 1568239201^{5}$

$2003^{15}=2003^{3\times5}=(2003^{3})^5 =8036054027^{5}$

$1568239201<8036054027=>199^{20}<2003^{15}$

d, $3^99 =3^{3\times33}=(3^{3})^{33}=27^{33}>27^{21}$

$11^{21}<27^{21}=>3^{99}>11^{21}$

$3^{2n}=9^n$

$2^{3n}=8^n$

$9>8=>3^{2n}>2^{3n}$

So sánh các số sau

a) 5³ và 3⁵

5³ = 125

3⁵ = 243

=> 5³ < 3⁵

3² và 2³

3² = 9

2³ = 8

=> 3² > 2³

2⁶ và 6²

2⁶ = 64

6² = 36

=> 2⁶ > 6²

b) 2015 x 2017 và 2016²

2015 x 2017 

= 2015 x ( 2016 + 1 ) 

= 2015 x 2016 + 2015 

2016²

= 2016 x 2016

= 2016 x ( 2015 + 1 )

= 2016 x 2015 + 2016

Vì: 2015 < 2016

=> 2015 x 2017 < 2016²

c) 199²⁰ và 2003¹⁵

199²⁰ < 200²⁰ = ( 2³ x 5² )²⁰ = 2⁶⁰ x 5⁴⁰

2003¹⁵ > 2000¹⁵ = ( 2 x 10³ )¹⁵ = ( 2⁴ x 5³ )¹⁵ = 2⁶⁰ x 5⁴⁵

=> 2003¹⁵ > 199²⁰

d) 3⁹⁹ và 11²¹

3⁹⁹ = ( 3³ )³³ = 27³³ > 27²¹

Vì: 27 > 11

=> 27²¹ > 11²¹ 

=> 3⁹⁹ > 11²¹

3²ⁿ và 2³ⁿ

3²ⁿ = ( 3² )ⁿ = 9ⁿ

2³ⁿ = ( 2³ )ⁿ = 8ⁿ

Vì 9 > 8

=> 9ⁿ > 8ⁿ

=> 3²ⁿ > 2³ⁿ

Vậy 3²ⁿ > 2³ⁿ

a) 5³⁶ và 11²⁴

Ta có: 5³⁶= (5³)¹²=125¹²  (1)

             11²⁴=(11²)¹²=121¹² (2)

Từ (1) và (2) => 5³⁶>11²⁴

^{tương tự.....}

Đáp án là :

câu 20 :625 < 1257

câu 21 :536 > 1124

câu 22 :32n < 23n

câu 23 :523 < 6.522

câu 24 :1124 <19920

câu 25 :399 > 112

a) Ta có: \(5^{36}=\left(5^3\right)^{12}=125^{12}\)

\(11^{24}=\left(11^2\right)^{12}=121^{12}\)

mà \(125^{12}>121^{12}\left(125>121\right)\)

nên \(5^{36}>11^{24}\)

b) Ta có: \(3^{2n}=\left(3^2\right)^n=9^n\)

\(2^{3n}=\left(2^3\right)^n=8^n\)

mà \(9^n>8^n\left(9>8\right)\)

nên \(3^{2n}>2^{3n}\)

a: 199^20=1568239201^5

2003^15=8036054027^5

=>199^20<2003^15

b: 3^99=27^33>27^21=11^21

Lời giải:

a. 

$199^{20}<200^{20}=(2.100)^{20}=2^{20}.10^{40}=(2^{10})^2.10^{40}< (10^4)^2.10^{40}=10^8.10^{40}=10^{48}$
$2003^{15}> 2000^{15}=(2.10^3)^{15}=2^{15}.10^{45}> 2^{10}.10^{45}> 10^3.10^{45}=10^{48}$

$\Rightarrow 199^{20}< 2003^{15}$
b.

$3^{99}=(3^9)^{11}=19683^{11}$
$11^{21}< 11^{22}=(11^2)^{11}=121^{11}$
Hiển nhiên $19683^{11}> 121^{11}$

$\Rightarrow 3^{99}> 121^{11}> 11^{21}$

a, Ta có : \(8>7\)

\(\Rightarrow2^{13}.8=2^{16}>2^{13}.7\)

b, Ta có : \(199^{20}< 200^{20}=2^{60}.5^{40}\)

Mà \(2003^{15}>2000^{15}=2^{60}.2^{45}\)

Thấy : \(45>40\)

\(\Rightarrow2000^{15}>200^{20}\)

\(\Rightarrow2003^{15}>199^{20}\)

c, Ta có : \(\left\{{}\begin{matrix}202^{303}=\left(2.101\right)^{3.101}=\left(8.101^3\right)^{101}\\303^{202}=\left(3.101\right)^{2.101}=\left(9.101^2\right)^{101}\end{matrix}\right.\)

Mà \(8.101^3>9.101^2\)

\(\Rightarrow202^{303}>303^{202}\)

a) Ta có: \(2^{16}=2^{13}\cdot8\)

mà \(7< 8\)

nên \(7\cdot2^{13}< 2^{16}\)

b) \(199^{20}=1568239201^5\)

\(2003^{15}=8036054027^5\)

mà \(1568239201< 8036054027\)

nên \(199^{20}< 2003^{15}\)

c) Ta có: \(202^{303}=\left(202^3\right)^{101}\)

\(303^{202}=\left(303^2\right)^{101}\)

mà \(202^3>303^2\)

nên \(202^{303}>303^{202}\)

Lo học đi bạn!

19 tháng 8 năm 2001

\(19920-75-85+35+6\)

\(=19845-85+35+6\)

\(=19801\)

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