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) \(100-\left(3.5^2-2.3^3\right)\)

\(100-\left(75-52\right)=100-23=77\)

b)\(4.5^2-3.2^3+3^9:3^7\)

\(100-24+9=85\)

c) \(\left(392:7\right).2^3-2^3+2020^0\)

\(8\left(56-1\right)+1=441\)

d) \(\left(6^2+7^2+8^2+9^2+10^2\right)-\left(1^2+2^2+3^2+4^2+5^2\right)=275\)

(lười - thông cảm :(

Đây bạn ơi a,\(100-\left(3.5^2-2.3^3\right)=100-\left(75-54\right)\))=100-19=81

b,\(4.5^2-3.2^3+\frac{3^9}{3^7}=100-24+9=85\)

c,\(\left(392:7\right).2^3-2^3+2020^0=56.2^3-2^3+1=441\)

d,\(=\left(6^2-1^2\right)+\left(7^2-2^2\right)+\left(8^2-3^3\right)+\left(9^2-4^2\right)+\left(10^2-5^2\right)\)

=\(5.7+5.9+5.11+5.13+5.15=5\left(7+9+11+13+15\right)\)

=\(5.55=275\)

3 không chia hết cho 2 nên 

\(3^{5^7}\) không chia hết cho 2 

Vậy A = 1999^2k+1

      A = (1999²)^k.1999

    A = \(\overline{...1}\)^k.1999

    A = \(\overline{..9}\)

Vì 6 ⋮ 2 nên \(6^{8^9}\) ⋮ 2

Vậy B = 2024^2k = (2024²)^k = \(\overline{..6}\)^k = \(\overline{..6}\)

\(S=1+2+2^2+2^3+...+2^{2020}+2^{2021}\)

\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{2020}+2^{2021}\right)\)

\(=3+2^2\left(1+2\right)+...+2^{2020}\left(1+2\right)\)

\(=3+2^2.3+...+2^{2020}.3⋮3\)

     VẬY \(S⋮3\)

Trả lời :...........................................

SCSH: (2021 - 1) : 1 = 2020

Tổng: (2021 + 1) : 2 = 1011

Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

k nhé

\(S=1+2+2^2+2^3+...+2^{2020}+2^{2021}\)

\(\text{Số số hạng của S là 2022 số , chia làm 1011 cặp , mỗi cặp 2 số .}\)

\(\Leftrightarrow S=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{2020}+2^{2021}\right)\)

\(\Leftrightarrow S=3+2^2\left(1+2\right)+...+2^{2020}\left(1+2\right)\)

\(\Leftrightarrow S=3+2^2\times3+...+2^{2020}\times3\)

\(\Leftrightarrow S=3\left(1+2^2+...+2^{2020}\right)\)

\(\Rightarrow S⋮3\left(đpcm\right)\)

a) Vì \(-45< -16\) nên \(\left(-\dfrac{45}{17}\right)^{15}< \left(\dfrac{-16}{17}\right)^{15}\)

b) Vì \(21< 23\) nên \(\left(-\dfrac{8}{9}\right)^{21}< \left(-\dfrac{8}{9}\right)^{23}\)

c) \(27^{40}=3^{3^{40}}=3^{120}\)

\(64^{60}=8^{2^{60}}=8^{120}\)

Vì \(3< 8\) nên \(3^{120}< 8^{120}\) hay \(27^{40}< 64^{60}\)

con ai kooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

co don quaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

2+3= mấy

2+3=5

vậy thì đưa bàn tay của e cho a nắm này