\(x+2=x+1+1⋮\left(x+1\right)\Leftrightarrow1⋮\left(x+1\right)\)

\(\Leftrightarrow x+1\inƯ\left(1\right)=\left\{-1,1\right\}\Leftrightarrow x\in\left\{-2,0\right\}\).

\(3^1+3^2+3^3+3^4+...+3^{2012}\)

\(=\left(3^1+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)

\(=\left(3^1+3^2+3^3+3^4\right)+3^4\left(3^1+3^2+3^3+3^4\right)+...+3^{2008}\left(3^1+3^2+3^3+3^4\right)\)

\(=120\left(1+3^4+...+3^{2008}\right)\)chia hết cho \(120\).

760 triệu cả năm nhé bạn

k cho mik với

760 TR

         4^x + 4^x+1=320
4^x x 1 + 4^x x 4 =320

4^x x ( 1 + 4 )   =320

4^x x 5              =320

4^x                      =320 : 5

4^x                      = 64

4^x                   = 4³

=>            x    = 3

Vậy x = 3

Bài 5: 

\(6a+1=6a-2+3=2\left(3a-1\right)+3⋮\left(3a-1\right)\Leftrightarrow3⋮\left(3a-1\right)\)

\(\Leftrightarrow3a-1\inƯ\left(3\right)=\left\{-3,-1,1,3\right\}\Leftrightarrow a=0\)(vì \(a\)là số nguyên) 

mk làm phần b bài 4 nha:

Ta có A=[1-(-1)+(-2)]-[1-(-1)-(-2)]

         A=[0+(-2)]+[0-(-2)]

         A=0-2+0+2

         A=(2-2)+(0+0)

         A=0   +    0

         A=0

Vậy A=0

=D

TL:

A = { x € Z/-2<x<5}

A = { -1 ; 0 ; 1 ; 2 ; 3 ; 4}

HT

TL:

(-2)-5=(-2)+(-5)=-7

HT