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 \(202^{303}>303^{202}\)

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

sorry nghe h tớ gửi quá 100 tin nhắn nên nó ko cho gửi

Bài 1

a)2711>818

b)6255>1257

c)536<1124

d)32n>23n

Bài 2

a)523<6.522

b)7.213>216

c)2115<275.498

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) 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

199^20 < 2003^15

199²⁰<2003¹⁵

2115 = (7.3)15 = 715.315

275.498 = (33)5.(72)8 = 315.716 = 7.315.715 > 315.715 = 2115

=> 275.498 > 2115.

tick cho mink nhé 😊

a)Ta có:  \(5^{36}=5^{3.12}=\left(5^3\right)^{12}=125^{12}\)

              \(11^{24}=11^{2.12}=\left(11^2\right)^{12}=121^{12}\)

Vì \(125>121\Rightarrow125^{12}>121^{12}\)

\(\Rightarrow5^{36}>11^{24}\)

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

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

Vì \(20< 21\Rightarrow5^{20}< 5^{21}\)

\(\Rightarrow625^5< 125^7\)

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

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

Vì \(9>8\Rightarrow9^n>8^n\)( do \(n>0\))

\(\Rightarrow3^{2n}>2^{3n}\)

d)Ta có:  \(5^{23}=5.5^{22}< 6.5^{22}\)

\(\Rightarrow5^{23}< 6.5^{22}\)

a. 5^36=(5^3)^12

               =125^12

11^24=(11^2)^12

         = 121^12

Vì 125^12>121^12 nên 5^36>11^24

b. Ta có: 625^5 =(5^4)^5

                         = 5^20

               125^7=(5^3)^7

                        = 5^21

 Vì 5^20<5^21 nên 625^5<125^7

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)15va 18  

15=3.5

18=3^2.2

BCNN(15;18)=2.3^2.5=90

d)33;44va55

33=3.11

44=4.11

55=5.11

BCNN(33;44;55)=3.4.5.11=660

c)4;14 va 26

4=2^2

14=2.7

26=2.13

BCNN(4;14;26)=2^2.7.13=364

\(a.15=3.5;18=2.3^2\)

\(BCNN\left(15;18\right)=2.3^2.5=90\)
 

\(b.33=3.11;44=2^2.11;55=5.11\)

\(BCNN\left(33;44;55\right)=2^2.3.5.11=660\)