\(33-6n=30+3-6n=30-3\left(2n-1\right)\)

Để \(33-6n⋮2n-1\) thì \(\left(2n-1\right)\inƯ\left(30\right)=\left\{1;2;3;5;6;10;15;30\right\}\)

Mà \(n\in N\Rightarrow n\in\left\{1;2;3;8\right\}\)

x = 4

y = 5

tính từng cái 1 ra rồi kết luận 

( x + 1 ) ^3 - 4 = 60

= ( x + 1 ) ^3 = 60 + 4

=  ( x + 1 ) ^3  = 64

=  ( x + 1 ) ^3  = 4^3

= x + 1 = 4

= x = 4 - 1

= x = 3 

\(2\times3^{x+1}-3^x=135\\ 2\times3\times3^x-3^x=135\\ 6\times3^x-3^x=135\\ 5\times3^x=135\\ 3^x=27\\ 3^x=3^3\\ x=3\)

Ta có \(2.3^{x+1}-3^x=135\)

\(\Leftrightarrow2.3^x.3-3^x=135\)

\(\Leftrightarrow3^x.6-3^x=135\)

\(\Leftrightarrow3^x\left(6-1\right)=135\)

\(\Leftrightarrow3^x=27\)

\(\Leftrightarrow3^3=27\Leftrightarrow x=3\)

Vậy x=3

A = 26²⁰¹¹ + 8²⁰²² + 2022²⁰²³

A = ( ...6 ) + 8^{2020 + 2} + 2022^{2020 + 3}

A = ( ...6 ) + 8^{505 * 4} + 8² + 2022^{505 * 4} + 2022³

A = ( ...6 ) + ( 8⁴ )⁵⁰⁵ + 64 + ( 2022⁴ )⁵⁰⁵ + ( ...8 )

A = ( ...6 ) + ( ...6 )⁵⁰⁵ + 64 + ( ...6 )⁵⁰⁵ + ( ...8 )

A = ( ...6 ) + ( ...6 ) + 64 + ( ...6 ) + ( ...8 )

A = ( ...0 )

Vậy A có chữ số tận cùng là 0

Ta có : P = 29 - [4²+3¹.(51-7²)] + 2021

=> P = 29 - [16+3.(51-49)] + 2021

=> P = 29-(16+3.3)+2021

=> P = 29 - ( 16 + 9 ) + 2021

=> P = 29 - 25 + 2021

=> P = 4 + 2021 = 2025

Ta có : P = 29 - [42+31.(51-72)] + 2021

=> P = 29 - [16+3.(51-49)] + 2021

=> P = 29-(16+3.3)+2021

=> P = 29 - ( 16 + 9 ) + 2021

=> P = 29 - 25 + 2021

=> P = 4 + 2021 = 2025

Vậy P = 2025

a) Ta có :

\(\dfrac{19}{40}=\dfrac{38}{80}< \dfrac{41}{80}\)

b) Ta có :

\(\dfrac{19}{26}< \dfrac{21}{26}< \dfrac{21}{25}\)

41/80 lớn hơn 19/40

21/25 lớn hơn 19/ 26