a) Ta có:

\(2^{300}=2^{3\cdot100}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=3^{2\cdot100}=\left(3^2\right)^{100}=9^{100}\)

Mà: \(8< 9\)

\(\Rightarrow8^{100}< 9^{100}\)

\(\Rightarrow2^{300}< 3^{200}\)

b) Ta có:

\(3^{500}=3^{5\cdot100}=\left(3^5\right)^{100}=243^{100}\)

\(7^{300}=7^{3\cdot100}=\left(7^3\right)^{100}=343^{100}\)

Mà: \(243< 343\)

\(\Rightarrow243^{100}< 343^{100}\)

\(\Rightarrow3^{500}< 7^{300}\)

c) Ta có: 

\(8^5=\left(2^3\right)^5=2^{3\cdot5}=2^{15}=2\cdot2^{15}\)

\(3\cdot4^7=3\cdot\left(2^2\right)^7=3\cdot2^{2\cdot7}=3\cdot2^{14}\)

Mà: \(2< 3\)

\(\Rightarrow2\cdot2^{14}< 3\cdot2^{14}\)

\(\Rightarrow8^5< 3\cdot4^7\)

d) Ta có:

\(202^{303}=202^{3\cdot101}=\left(202^3\right)^{101}=8242408^{101}\)

\(303^{202}=303^{2\cdot101}=\left(303^2\right)^{101}=91809^{101}\)

Mà: \(8242408>91809\)

\(\Rightarrow8242408^{101}>91809^{101}\)

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

a)\(27^2\)và \(4^6\)

\(27^2=\left(3^3\right)^2\)

\(4^6=\left(2^3\right)^2\)

\(3^3>2^3\)

b) \(3^{500}=\left(3^5\right)^{100}\)

\(7^{300}=\left(7^3\right)^{100}\)

\(7^3=343\)

\(3^5=243\)

\(\Rightarrow3^{500}< 7^{300}\)

c) \(8^5=4^5\cdot2^5\)

\(3\cdot4^7=3\cdot4^2\cdot4^5\)

\(3\cdot4^2>2^5\)

\(3\cdot4\cdot4=2\cdot2\cdot2\cdot2\cdot3>2\cdot2\cdot2\cdot2\cdot2\)

\(8^5< 3\cdot4^7\)

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

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

\(202^3>303^2\)

Nên

cam on ban

Câu 1.99²⁰và 9999¹⁰

⁼(99²)¹⁰=9801¹⁰

Vậy 9801¹⁰<9999¹⁰ suy ra  99²⁰<9999¹⁰

Câu 2. 3⁵⁰⁰và 7³⁰⁰

 3⁵⁰⁰=(3⁵)¹⁰⁰=243¹⁰⁰

7³⁰⁰=(7³)¹⁰⁰=343¹⁰⁰

Vậy 243¹⁰⁰<343¹⁰⁰ => 3⁵⁰⁰<7³⁰⁰

Bạn có thể ghi cho tiết đề bài và bạn muốn làm gì cho bài đó được không?

a/

\(2^{1050}=\left(2^2\right)^{525}=4^{525}< 5^{525}< 5^{540}\)

b/

\(2^{161}>2^{160}=\left(2^4\right)^{40}=16^{40}>13^{40}\)

c/

\(17^{14}>16^{14}=\left(2^4\right)^{14}=2^{56}>2^{55}=\left(2^5\right)^{11}=32^{11}>31^{11}\)

`@` `\text {Ans}`

`\downarrow`

\(202^{303}\text{ và }303^{202}\)

Ta có:

\(202^{303}=202^{3\cdot101}=\left(202^3\right)^{101}\)

\(303^{202}=303^{101\cdot2}=\left(303^2\right)^{101}\)

So sánh `202^3` và `303^2`, ta có:

`202^3 = (2*101)^3 = 2^3 * 101^3 = 8 * 101^3 = 8* 101^2 * 101 = 808*101^2`

`303^2 = (3*101)^2 = 3^2 * 101^2 = 9 * 101^2`

Vì `9 < 808 \Rightarrow 9*101^2 < 808*101^2`

`\Rightarrow`\(202^{303}>303^{202}\)

Vậy, \(202^{303}>303^{202}.\)

1+1=3

a, 2^300 < 3^200

b, 3^500 < 7^300

c, 8^5 > 3.4^7

d, 202^303 < 303^202 

A) Ta có : 2³⁰⁰=2^3.100=(2³)¹⁰⁰=8¹⁰⁰

3²⁰⁰=3^2.100=(3²)¹⁰⁰=9¹⁰⁰

Vì 8¹⁰⁰<9¹⁰⁰

Nên 2³⁰⁰<3²⁰⁰

B)Ta có: 3⁵⁰⁰ =3^5.100=(3⁵)¹⁰⁰=243¹⁰⁰

7³⁰⁰=7^3.100=(7³)¹⁰⁰=343¹⁰⁰

Vì 243¹⁰⁰<343

Nên 3⁵⁰⁰<7³⁰⁰

bằng mắt híp

a) 5¹⁴ = 5^2.7= (5²)⁷= 25⁷

3²¹= 3^3.7= (3³)⁷= 27⁷

Mà 25 < 27 

=> 5¹⁴<3²¹

b) 4³³= (4³)¹¹  = 64¹¹

3⁴⁴= (3⁴)¹¹ = 81¹¹

=> 4³³<3⁴⁴

c) 2¹⁶¹ > 2¹⁶⁰ = (2⁴)⁴⁰= 16⁴⁰

Mà 13⁴⁰< 16⁴⁰ < 2¹⁶¹

=> 13⁴⁰<2¹⁶¹

d) 17¹⁵= 17^3.5= (17³)⁵ = 4913⁵

31¹⁰= (31²)⁵= 961⁵

Mà 4913 > 961

=> 17¹⁵> 31¹⁰

e) 2²⁵⁵ = (2⁵)⁴⁵ = 32⁴⁵

3¹⁸⁰ = (3⁴)⁴⁵ = 81⁴⁵

=> 2²⁵⁵ < 3¹⁸⁰

g) 202³⁰³ = (202³)¹⁰¹ = (101³.2³)¹⁰¹= (101³.8)¹⁰¹= (101².8.101)¹⁰¹= (101².808)¹⁰¹

303²⁰² = (303²)¹⁰¹= (3².101²)¹⁰¹= (101².9) ¹⁰¹

Vì 101².9 < 101².808

=> 202³⁰³>303²⁰²

a) ta có: 7^10 < 7^14 = (7^2)^7 = 49^7 < 50^7

=> 7^10 < 50^7

b) ta có: 5^30 = (5^3)^10 = 125^10 > 124^10

=> 5^30 > 124^10

c) ta có: 9^21 = (9^3)^7=729^7

phần d thì mk ko bk, xl bn nha

a,<

b,>

c,=

^_^