M= ( 1+2010¹)+(2010²+2010³)+(2010⁴+2010⁵)+(2010⁶+2010⁷)

M= 1.(2010+1) + 2012².(2010+1)+2010⁴.(2010+1)+2010⁶.(2010+1)

M= 2011.(1+2012²+2010⁴+2010⁶)

Vậy M chia hết cho 2011

vì 2+2²=2+4=6 chia hết cho 3 nên 2+2²+2³+2⁴+2⁵+...+2⁶⁰chia hết cho 3

nên tổng A chia hết cho 3

vậy A chia hết cho 3

                                                          Giải :

A=(2 + 2^2) + ( 2^4 + 2^5 ) +........................ +(2^59 + 2^60) 

A = 2* ( 1 + 2 ) + 2^4* (1+2)+.........................+2^59 *(1+2)

Ta có 1+2=3 

Mà 3 chia hết cho 3 nên A chia hết cho 3

[x+1]+[x+2]+...+[x+100]=5750

[1+2+...+100].x=5750

Số các số hạng:

(100-1):1+1=100 số

=>có 100 thừa số x

Tổng trên là:

(1+100)x100:2=5050

=>x.100=5750-5050

=>x.100=700

=>x=700:100

=>x=7

Vậy x = 7

Ta có: (x + 1) + (x + 2) +...+ (x + 100) = 5750

=> (x + x + x +...... +x) + (1 + 2 + 3 + 4 + ... + 100) = 5750    

=> 100x + 5050 = 5750

=> 100x = 5750 - 5050

=> 100x = 700

=> x = 700 : 100

=> x = 7

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uytyyyy

1;2;4;5;7;8;................

goi ucln (4n+3,2n+3) la d(d thuoc N*) 

<=>4n+3 chia het cho d,2n+3 chia het cho d

<=>2.(2n+3)-4n+3

<=>3 chia het cho d <=>d thuoc tap hop {1;3}

do 4n va 2n chan =>2n+3 va 4n+3 ko chia het cho3

=>d=1

<=>n thuoc tap hop 1,2