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: 5^23=5*5^22<6*5^22

=>6*5^22 lớn hơn

b: 7<8

=>7*2^13<8*2^13=2^16

=>2^16 lớn hơn

c: 21^15=3^15*7^15

27^5*49^8=3^15*7^16

mà 15<16

nên 27^5*49^8 lớn hơn

a) Ta có:

5²³ = 5.5²²

Do 6 > 5 nên 6.5²² > 5.5²²

Vậy 6.5²² > 5²³

b) Ta có:

2¹⁶ = 2³.2¹³ = 8.2¹³

Do 8 > 7 nên 8.2¹³ > 7.2¹³

Vậy 2¹⁶ > 7.2¹³

c) Ta có:

21¹⁵ = (3.7)¹⁵ = 3¹⁵.7¹⁵

27⁵.49⁸ = (3³)⁵.(7²)⁸ = 3¹⁵.7¹⁶

Do 16 > 15 nên 7¹⁶ > 7¹⁵

⇒ 3¹⁵.7¹⁶ > 3¹⁵.7¹⁵

Vậy 27⁵.49⁸ > 21¹⁵

72^45-72^44=72^44(72-1)=72^44*71

72^44-72^43=72^43(72-1)=72^43*71

=>72^45-72^44>72^44-72^43

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 \(202^{303}>303^{202}\)

199^20 < 2003^15

199²⁰<2003¹⁵

<

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 

\(a.10^{30}=\left(10^3\right)^{10}=1000^{10}\\ 2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)

Vì 1000¹⁰ < 1024¹⁰ => 10³⁰ < 2¹⁰⁰

\(b,333^{444}=\left(111\cdot3\right)^{444}=111^{444}\cdot3^{444}=111^{444}\cdot81^{111}\\ 444^{333}=\left(111\cdot4\right)^{333}=111^{333}\cdot4^{333}=111^{333}\cdot64^{111}\)

Vì 111⁴⁴⁴ >111³³³ ; 81¹¹¹ > 64¹¹¹ => 333⁴⁴⁴ > 444³³³

mình cảm ơn

Ta có:

$3^{39}=3^{3\times33}=(3^{3})^{33}=27^{33}>27^{21}$

Mà $11^{21}<27^{21}=>3^{39}>11^{21}$

3³⁹   = (3¹³)³

11²¹ = (11⁷)³

3¹³  = (3²)⁶.3 = 9⁶.3  < 11⁶. 11 = 11⁷

⇒ 3¹³ < 11⁷ ⇒ (3¹³)³ < (11⁷)³

⇒ 3³⁹ < 11²¹