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) \(243^5=\left(3^5\right)^5=3^{25}\)

\(3\cdot27^5=3\cdot\left(3^3\right)^5=3\cdot3^{15}=3^{16}\)

mà \(3^{25}>3^{16}\)

nên \(243^5>3\cdot27^5\)

b) \(625^5=\left(5^4\right)^5=5^{20}\)

\(125^7=\left(5^3\right)^7=5^{21}\)

mà \(5^{20}< 5^{21}\)

nên \(625^5< 125^7\)

c) \(202^{303}=\left(202^3\right)^{101}=8242408^{101}\)

\(303^{202}=\left(303^2\right)^{101}=91809^{101}\)

mà \(8242408^{101}>91809^{101}\)

nên \(202^{303}>303^{202}\)

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 

còn cách nào ra số nhỏ hơn ko bạn

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

199^20 < 2003^15

199²⁰<2003¹⁵

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

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

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

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

\(3^2;2^3\)

\(=9;8\)

\(=3^2>2^3\)

Có đúng ko bạn?

3 mũ 2 thì =9

2 mũ 3 = 8 

mà 9>9

-> 3mu2> 2 mũ 3

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 

3 mũ 99  <  11 mũ 21

mình nghĩ    11 mũ 21 < 3 mũ 99