Bài 1:

\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)

\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)

Bài 2:

\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)

đúng rùi

Đặt S=1+3+3²+3³+...+3⁵⁰

3S=3+3²+3³+...+3⁵¹

3S-S=3-3+3²-3²+..3⁵⁰-3⁵⁰+3⁵¹-1

2S=3⁵¹-1

S=(3⁵¹-1) :2

nhân 3 cả vế lên rồi trừ cho vế trước sau đó chia 2 thì ra

a) Vẽ đường thẳng D sao cho Mlà trung điểm AD

Xét tam giác AMB và DMC:

AM=ME

Góc AMB=góc CME

MB=MC

=> Tam giác AMB=DMC (Cạnh.góc.cạnh)

=> AB=CD; góc BAM=góc D

Ta có; AC>AB nên AB=CD

=> Tam giác ACD có AC=CD

=> góc D= góc MAC

<=> góc BAM> CAM (đpcm)

bạn ơi còn hai câu cuối thì làm thế nào

\(A=1+2+2^2+2^3+...+2^{119}\)

\(2A=2+2^2+2^3+...+2^{120}\)

\(2A-A=\left(2+2^2+2^3+...+2^{120}\right)-\left(1+2+2^2+2^3+...+2^{119}\right)\)

\(A=2^{120}-1\)

Có \(120\)chia hết cho các số \(2,3,8,5\)nên \(A\)chia hết cho \(2^2-1=3,2^3-1=7,2^8-1=255=17.15,2^5-1=31\).

Suy ra đpcm. 

\(A=1+2^1+2^2+...+2^{100}+2^{101}\)

\(=\left(1+2^1+2^2\right)+\left(2^3+2^4+2^5\right)+...+\left(2^{99}+2^{100}+2^{101}\right)\)

\(=\left(1+2^1+2^2\right)+2^3\left(1+2^1+2^2\right)+...+2^{99}\left(1+2^1+2^2\right)\)

\(=7\left(1+2^3+...+2^{99}\right)\)chia hết cho \(7\).