\(a)\) Ta có: \(99^{20}=\left(99^2\right)^{10}=9801^{\text{ }10}\)

Vì \(9801< 9999\)nên \(9801^{10}< 9999^{10}\)

Hay \(99^{20}< 9999^{10}\)

Vậy 99²⁰<9999¹⁰

ta có:

1/10.A=10¹⁰⁰+1/10(10⁹⁹+1)

1/10.A=10¹⁰⁰+1/10¹⁰⁰+10

1/10.A=1-(9/10¹⁰⁰+10)

1/10.B=10¹⁰¹+1/10(10¹⁰⁰+1)

1/10.B=10¹⁰¹+1/10¹⁰¹+10

1/10.B=1-(9/10¹⁰¹+10)

vì(10¹⁰¹+10)>(10¹⁰⁰+1)=>  9/10¹⁰¹+10 < 9/10¹⁰⁰+10 => 1-(9/10¹⁰¹+10) > 1-(9/10¹⁰⁰+10)

hay 1/10.A>1/10.B

=>A>B

ta có:

1/10.A=10100+1/10(1099+1)

1/10.A=10100+1/10100+10

1/10.A=1-(9/10100+10)

1/10.B=10101+1/10(10100+1)

1/10.B=10101+1/10101+10

1/10.B=1-(9/10101+10)

vì(10101+10)>(10100+1)=>  9/10101+10 < 9/10100+10 => 1-(9/10101+10) < 1-(9/10100+10)

hay 1/10.A<1/10.B

=>A<B

Đáp án dưới mới đúng nhé 

vừa mình làm nhầm

c: \(100C=\dfrac{100^{100}+100}{100^{100}+1}=1+\dfrac{99}{100^{100}+1}\)

\(100D=\dfrac{100^{101}+100}{100^{101}+1}=1+\dfrac{99}{100^{101}+1}\)

100^100+1<100^101+1

=>\(\dfrac{99}{100^{100}+1}>\dfrac{99}{100^{101}+1}\)

=>100C>100D

=>C>D

b: \(2020E=\dfrac{2020^{2022}+2020}{2020^{2022}+1}=1+\dfrac{2019}{2020^{2022}+1}\)

\(2020F=\dfrac{2020^{2021}+2020}{2020^{2021}+1}=1+\dfrac{2019}{2020^{2021}+1}\)

2020^2022+1>2020^2021+1(Do 2022>2021)

=>\(\dfrac{2019}{2020^{2022}+1}< \dfrac{2019}{2020^{2021}+1}\)

=>2020E<2020F

=>E<F

5^255 để nguyên, còn 2^572 tách ra thành 4^255 . 2^62 
bấm máy căn bậc 255 của 2^62 rồi nhân với 4, sau đó so sánh với 5 =>ok 
Con thứ 2: 
8^9+7^9+6^9 +.... + 1^9 < 8 . 8^9 = 8^10 
Mà 8^10 < 9^10 => 9^10 > 8^9 + 7^9 + 6^9 +...+ 1^9

k cho mình nha

Viết rối qá chả thấy j.

\(99^2vs9999^{10}\)

\(9999^{10}=\left(101\cdot99\right)^{10}=101^{10}\cdot99^{10}\)

Vì \(99^{10}>99^2=>99^2< 9999^{10}\)

a) Ta có: 2^91 = (2^13)^7 = 8192^7

5^35 = (5^5)^7 = 3125^7

Vì 8192 > 3125 nên 8192^7 > 3125^7

Vậy 2^91 > 5^35

b) Ta có: 9999^10 = 99^10 . 101^10

Vì 99^2 < 99^10 nên 99^2 < 99^10 . 101^10

Vậy 99^2 < 9999^10

c) Ta có: 2^300 = (2^6)^50 = 64^50

3^200 = (3^4)^50 = 81^50

Vì 49 < 64 < 81 nên 49^50 < 64^50 < 81^50

Vậy 49^50 < 2^300 < 3^200

d) 9^3/25^3 = (9/25)^3

3^6/2^12 = (3^2)^3/(2^4)^3 = 9^3/16^3 = (9/16)^3

Vì 9/25 < 9/16 nên (9/25)^3 < (9/16)^3

Vậy 9^3/25^3 < 3^6/2^12.

Ta có:

\(5^{225}=5^{15.17}=\left(5^{15}\right)^{17}\)=30517578125¹⁷
2^572=2^16.17=(2^16)^17=65536^17
=>5^255 > 2^572

5^4 = 625 ; 2^9 = 512 
=> 5^4 > 2^9

=> (5^4)^64 > (2^9)^64

=> 5^255 > 2^576 / 5 > 2^576 / 2^3 = 2^573 > 2^572 
Vậy 5^255 > 2^572

Ta thấy: 2⁹=512 và 5⁴=625
=> 2⁹ <5⁴ => (2⁹)⁶⁴ <(5⁴)⁶⁴
=> 2⁵⁷⁶ < 5²⁵⁶
Mà 5²⁵⁵ < 5²⁵⁶ ; 2⁵⁷² < 2⁵⁷⁶
=> 5²⁵⁵ > 2⁵⁷²