(9x + 5).4 - 20 = 420

(9x + 5).4 =440

9x + 5=110

9x=105

x=35/3

Bạn ơi chia hết cho 3 rồi còn bé hơn số gì nữa không?

a∈A  

b∈A

u∈B

a∉B

b∉B

u∉A

\(\hept{\begin{cases}a\in A\\a\notin B\end{cases}}\) \(\hept{\begin{cases}b\in A\\b\in B\end{cases}}\) \(\hept{\begin{cases}u\notin A\\u\in B\end{cases}}\)

ta có :

\(B=\left[\left(1+3^4\right)+\left(3+3^5\right)+\left(3^2+3^6\right)+\left(3^3+3^7\right)\right]+\left[..\right]..+\left[...+\left(3^{2010}+3^{2014}\right)+\left(3^{2011}+3^{2015}\right)\right]\)

\(=82\left(1+3+3^2+3^3\right)+82\left(3^8+3^9+3^{10}+3^{11}\right)+..+82\left(3^{2008}+2^{2009}+3^{2010}+3^{2011}\right)\)

chia hết cho 82 nên cũng chia hết cho 41

52** : 33 = *6*

5280 : 33 = 161 nha 

           HOK TỐT !!

5280 : 33 = 160

k cho mik nha

ta có :

\(a.\hept{\begin{cases}18=2.3^2\\24=2^3.3\\54=2.3^3\end{cases}\Rightarrow x=2^3.3^3=216}\)

\(b.\hept{\begin{cases}84=2^2.3.7\\36=2^2.3^2\end{cases}\Rightarrow x=2^2.3=12}\)

Giải :

a) x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 50 )  = 2040

=> 51x + ( 1 + 2 + 3 + ... + 50 ) = 2040

=> 51x + ( 50 + 1 ) . 50 : 2 = 2040

=> 51x + 1275 = 2040

=> 51x = 765

=> x = 15

b) 1 + 2 + 3 + ... + x = 300

=> ( x + 1 ) . x : 2 = 300

=> ( x + 1 ) . x = 600

=> ( x + 1 ) . x = 25 . 24

=> x = 24

\(\left(x+1\right).\left(2y-5\right)=143\)

\(\Rightarrow\)\(\left(2y-5\right)\inƯ\left(143\right)\)

Mà 2y - 5 lẻ

\(\Rightarrow\)\(2y-5\in\left\{-1;-11;1;11\right\}\Rightarrow y=\left\{2;-3;3;8\right\}\)

Mà \(y\in N\Rightarrow y=\left\{2;3;8\right\}\)

\(\Rightarrow\)\(x+1\in\left\{-143;143;13\right\}\)

\(\Rightarrow\)\(x\in\left\{-144;142;12\right\}\)

Mà \(x\in N\Rightarrow x\in\left\{142;12\right\}\)

Vậy \(\left(x;y\right)=\left(142;3\right);\left(12;8\right)\)

ta có : \(\left(x+1\right)\left(2y-5\right)=13\times11\)

ta sẽ có 4 trường hợp nhau sau : \(\hept{\begin{cases}x+1=1\\2y-5=143\end{cases}\text{ hoặc }\hept{\begin{cases}x+1=11\\2y-5=13\end{cases}\text{ hoặc }}}\hept{\begin{cases}x+1=13\\2y-5=11\end{cases}\text{ hoặc }}\hept{\begin{cases}x+1=143\\2y-5=1\end{cases}}\)

ta giải ra được \(\left(x,y\right)\in\left\{\left(0,74\right);\left(10,9\right);\left(12,8\right);\left(142,3\right)\right\}\)

Bạn Đúc giúp người kiểu giì đấy :))) , giúp mà không giúp hết à ???

a) 2^x + 2020  2021

=> 2^x = 2021 - 2020

=> 2^x = 1

=> 2^x = 2⁰

=> x = 0

b) Ta có :

4x + 14 ⋮ x + 2

=> 4. ( x + 2 ) + 6 ⋮ x + 2

Mà 4 . ( x + 2 ) ⋮ x + 2 

=> 6 ⋮ x + 2 => x + 2 ∈ { 1 ; 2 ; 3 ;6 }

=> x ∈ { 0 ; 1 ; 4 } ( do x ∈ N )

c) ( x - 3 )²⁰²¹ - ( x - 3 )⁵ = 0

=> ( x - 3 )⁵ . [ ( 2 - 3 )²⁰¹⁶ - 1 ] = 0

\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^5=0\\\left(x-3\right)^{2016}-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^{2016}=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x-3\in=\left\{-1;1\right\}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x\in=\left\{2;4\right\}\end{cases}}\)

a) 2^x = 2021 - 2020

    2^x = 1

\(\Rightarrow\)2^x = 1⁰

\(\Rightarrow\)x = 0