P =2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10  

    = ( 2+2^2 ) + ( 2^3 + 2^4 ) + ( 2^5 + 2^6 ) + ( 2^7 + 2^8 ) + ( 2^9 + 2^10 )

   = 2.( 1+2 ) + 2^3.( 1+2 ) + 2^5.( 1+2 ) + 2^7.( 1+2 ) + 2^9.( 1 + 2 )

   = 2.3+2^3.3+2^5.3+2^7.3+2^9.3

   = ( 2+2^3+2^5+2^7+2^9).3 chia hết cho 3

=> P chia hết cho 3

câu 1 :19

câu 2:1

câu 3:3

câu 4:4

câu 5:có chia hết cho 3 vì tổng =2046

câu 1:19

câu 2:1

câu 3:3

câu 4:4

câu 5: có chia hết cho ba vì tổng = 2046

\(B=1+2+2^2+...+2^6.\)

\(=>4B=2^2+2^3+...+2^8\)\(\left(1\right)\)

\(A=2^2+2^3+...+2^8\)\(\left(2\right)\)

Từ (1) và (2) 

=> A = 4B

A = \(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+....+\frac{1}{2^{99}}\)

2²A = \(2+\frac{1}{2}+\frac{1}{2^3}+.....+\frac{1}{2^{97}}\)

=> 3A = \(2-\frac{1}{2^{99}}\)

=> A = \(\frac{2-\frac{1}{2^{99}}}{3}\)

a. \(A=2+2^2+2^3+2^4+2^5+...+2^{200}\)

\(2A=2^2+2^3+2^4+2^5+2^6+...+2^{201}\)

\(2A-A=2^{201}-2\)

b. \(Tacó:2^1=2;2^2=4;2^3=8;2^4=..6;2^5=2;2^6=4;...\)

Cứ dựa theo quy luật trên, ta có:

\(2^{4k+1}=..2;2^{4k+2}=...4;2^{4k+3}=...8;2^{4k}=..6\)

\(=>2^{201}=2^{4.50+1}\)

\(=>2^{101}=...2\)

\(=>2^{201}-2=..2-2\)\(=0\)

Vậy A có tân cùng là 0

ai nhanhh mk h

Bài 1: \(5^3+3^5\)

=125+243

=368

Bài 2:

a: \(\left(x-1\right)^3=125\)

=>\(\left(x-1\right)^3=5^3\)

=>x-1=5

=>x=6

b: \(\frac{720}{41-\left(2x-5\right)}=2^3\cdot5\)

=>\(\frac{720}{41-\left(2x-5\right)}=8\cdot5=40\)

=>41-(2x-5)=720/40=18

=>2x-5=23

=>2x=28

=>x=14

Bài 3:

a: \(\frac19\cdot3^4\cdot3^{n}=3^7\)

=>\(3^{n+4-2}=3^7\)

=>n+4-2=7

=>n+2=7

=>n=5

b: \(\left(2^2:4\right)\cdot2^{n}=4\)

=>\(\frac44\cdot2^{n}=4\)

=>\(2^{n}=4=2^2\)

=>n=2

B= ( 2 + 2^2 + 2^3 + 2^4 + 2^5) + 2^5. ( 2 + 2^2 + 2^3 + 2^4 + 2^5)+....+ 2^95 ( 2 + 2^2 + 2^3 + 2^4 + 2^5)

  = 62.(1 + 2^5 + ... + 2^95 ) chia hết cho 62

Suy ra B chia hết cho 31

 CAM ON NHE DUONG