1.

a.\(A=1+2^1+2^2+2^3+...+2^{2007}\)

\(2A=2+2^2+2^3+....+2^{2008}\)

b. \(A=\left(2+2^2+2^3+...+2^{2008}\right)-\left(1+2^1+2^2+..+2^{2007}\right)\)

\(=2^{2008}-1\) (bạn xem lại đề)

2.

\(A=1+3+3^1+3^2+...+3^7\)

a. \(2A=2+2.3+2.3^2+...+2.3^7\)

b.\(3A=3+3^2+3^3+...+3^8\)

\(2A=3^8-1\)

\(=>A=\dfrac{2^8-1}{2}\)

3

.\(B=1+3+3^2+..+3^{2006}\)

a. \(3B=3+3^2+3^3+...+3^{2007}\)

b. \(3B-B=2^{2007}-1\)

\(B=\dfrac{2^{2007}-1}{2}\)

4.

Sửa: \(C=1+4+4^2+4^3+4^4+4^5+4^6\)

a.\(4C=4+4^2+4^3+4^4+4^5+4^6+4^7\)

b.\(4C-C=4^7-1\)

\(C=\dfrac{4^7-1}{3}\)

5.

\(S=1+2+2^2+2^3+...+2^{2017}\)

\(2S=2+2^2+2^3+2^4+...+2^{2018}\)

\(S=2^{2018}-1\)

4:

a:Sửa đề: C=1+4+4^2+4^3+4^4+4^5+4^6

=>4*C=4+4^2+...+4^7

b: 4*C=4+4^2+...+4^7

C=1+4+...+4^6

=>3C=4^7-1

=>\(C=\dfrac{4^7-1}{3}\)

5:

2S=2+2^2+2^3+...+2^2018

=>2S-S=2^2018-1

=>S=2^2018-1

b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử  \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)

\(B\text{=}2.63+...+2^{56}.63\)

\(\Rightarrow B⋮63\)

\(\Rightarrow B⋮21\)

Bài 1

S₂ = 21 + 23 + 25 + ... + 1001

Số số hạng của S₂:

(1001 - 21) : 2 + 1 = 491

⇒ S₂  = (1001 + 21) . 491 : 2 = 250901

--------

S₄  = 15 + 25 + 35 + ... + 115

Số số hạng của S₄:

(115 - 15) : 10 + 1 = 11

⇒ S₄ = (115 + 15) . 11 : 2 = 715

Bài 2

a) 2x - 138 = 2³.3²

2x - 138 = 8.9

2x - 138 = 72

2x = 72 + 138

2x = 210

x = 210 : 2

x = 105

b) 5.(x + 35) = 515

x + 35 = 515 : 5

x + 35 = 103

x = 103 - 35

x = 78

c) 814 - (x - 305) = 712

x - 305 = 814 - 712

x - 305 = 102

x = 102 + 305

x = 407

d) 20 - [7.(x - 3) + 4] = 2

7(x - 3) + 4 = 20 - 2

7(x - 3) + 4 = 18

7(x - 3) = 18 - 4

7(x - 3) = 14

x - 3 = 14 : 7

x - 3 = 2

x = 2 + 3

x = 5

e) 9ˣ⁻¹ = 9

x - 1 = 1

x = 1 + 1

x = 2

2:

a: \(2x-138=2^3\cdot3^2\)

=>\(2x-138=8\cdot9=72\)

=>2x=138+72=210

=>x=105

b: \(5\cdot\left(x+35\right)=515\)

=>x+35=103

=>x=103-35=68

c: \(814-\left(x-305\right)=712\)

=>x-305=814-712=102

=>x=102+305=407

d: \(20-\left[7\left(x-3\right)+4\right]=2\)

=>7(x-3)+4=18

=>7(x-3)=14

=>x-3=2

=>x=5

e: \(9^{x-1}=9\)

=>x-1=1

=>x=2

f: \(5^{x-2}-3^2=2^4-\left(2^8\cdot2^2-2^{10}\cdot2^2\right)\)

=>\(5^{x-2}-9=16-1024+4096\)

=>\(5^{x-2}=3097\)

=>\(x-2=log_53097\)

=>\(x=2+log_53097\)

a: =(21-11)+(22-12)+(23-13)+(24-14)

=10+10+10+10

=40

b: =(34-14)+(35-15)+(36-16)+(37-17)

=20+20+20+20

=80

a, =(21-11)+(22-12)+(23-13)+(24-14)

=10+10+10+10

=40

b. =(34-14)+(35-15)+(36-16)+(37-17)

=20+20+20+20

=80

c,55+56+57+58-35-36-37-38

  =111+57+58-35-36-37-38

  =168+58-35-36-37-38

  =226-35-36-37-38

  =191-36-37-38

  =155-37-38

  =118-38

  =80

Số số hạng của biểu thức A là: (40-21):1+1=20(số hạng)

Ta có : 1/21>1/40,1/22>1/40,1/23>1/40,...,1/40=1/40

      1/21+1/22+1/23+...+1/40>1/40+1/40+1/41+1/40+...+1/40( 20 số 1/40)

      A>1/40x20=1/2

      A>1/20  (1)

Lại có: 1/21=1/21,1/21>1/22,1/21>1/23,...,1/21>1/40

      1/21+1/21+1/21+...+1/21(20 số 1/21)>1/21+1/22+1/23+...+1/40

      1/21x20>A

      20/21>A.Mà 1>20/21

    1>A   (2)

Từ (1) và (2) ta có : 1/2<A<1(đpcm)

Vậy bài tôán đđcm

\(\frac{1}{2}=\frac{1}{40}+\frac{1}{40}+....+\frac{1}{40}\)có 20 số hạng      \(\)

\(\frac{1}{21}+\frac{1}{22}+....+\frac{1}{40}\)có 20 số hạng

\(\frac{1}{21}>\frac{1}{40}\)

\(\frac{1}{22}>\frac{1}{40}\)

\(.....\)

\(\frac{1}{40}=\frac{1}{40}\)\(\Rightarrow\frac{1}{2}< \frac{1}{21}+\frac{1}{22}+.....+\frac{1}{40}\)

\(1=\frac{1}{40}+....+\frac{1}{40}\)có 40 số hạng mà A chỉ có 20 số hạng 

\(\Rightarrow\frac{1}{2}< A< 1\)

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

Easy!!

\(S=\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{35}>\dfrac{1}{29}+\dfrac{1}{29}+...+\dfrac{1}{29}\) (15 phân số \(\dfrac{1}{29}\))

\(=\dfrac{1.15}{29}=\dfrac{15}{29}>\dfrac{1}{2}\) (*)

\(\Rightarrow\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{35}>\dfrac{1}{2}^{\left(đpcm\right)}\)

P/s: đpcm là điều phải chứng minh

Có \(S=\dfrac{1}{21}+\dfrac{1}{22}+......+\dfrac{1}{35}\)

\(S=\dfrac{1}{21}+\dfrac{1}{22}+.........+\dfrac{1}{35}>\dfrac{1}{29}+\dfrac{1}{29}+\dfrac{1}{29}+........+\dfrac{1}{29}\)( 15 phân số \(\dfrac{1}{29}\))

\(S=\dfrac{15}{29}>\dfrac{1}{2}\)

\(S>\dfrac{1}{2}\)

Vậy S > \(\dfrac{1}{2}\)(đpcm)

1/20 .21 + 1/22 .23 + .... + 1/79 .80

= 1/20 - 1/21  + 1/22 - 1/23 + .......... + 1/79 - 1/80

= 1/20 - 1/80

= 3/80

Ta thấy : 3/80 < 1 

=> 1/20 . 21 + 1/22 . 23 + ........ + 1/79 . 80 <1 (ĐPCM)