1/

Tổng A là tổng các số hạng cách đều nhau 4 đơn vị.

Số số hạng: $(101-1):4+1=26$

$A=(101+1)\times 26:2=1326$

2/

$B=(1+2+2^2)+(2^3+2^4+2^5)+(2^6+2^7+2^8)+(2^9+2^{10}+2^{11})$

$=(1+2+2^2)+2^3(1+2+2^2)+2^6(1+2+2^2)+2^9(1+2+2^2)$

$=(1+2+2^2)(1+2^3+2^6+2^9)$

$=7(1+2^3+2^6+2^9)\vdots 7$

3/
$C=1+2+2^2+2^3+...+2^{99}$

$2C=2+2^2+2^3+2^4+...+2^{100}$

$\Rightarrow 2C-C=2^{100}-1$

$\Rightarrow C=2^{100}-1$

Sửa đề: A + 2 = 2^x-1

\(A=2+2^2+2^3+2^4+\dots+2^{10}\\2A=2^2+2^3+2^4+2^5+\dots+2^{11}\\2A-A=(2^2+2^3+2^4+2^5+\dots+2^{11})-(2+2^2+2^3+2^4+\dots+2^{10})\\A=2^{11}-2\\\Rightarrow A+2=2^{11}\)

Mà: \(A+2=2^{x-1}\)

\(\Rightarrow2^{x-1}=2^{11}\)

\(\Rightarrow x-1=11\)

\(\Rightarrow x=11+1=12\)

(1 + 2 + 22 + 23 + 24 + … + 210): 2047 
= [(1+210).210 : 2 ] : 2047
= [211. 105] : 2047
= 22155 : 2047
mình tính đến khúc này thì thấy chia ko hết :Đ
bạn xem lại đề hoặc có thể mik sai thật

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

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

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

=>\(x-2=log_53097\)

=>\(x=2+log_53097\)

\(\dfrac{2}{3}\)

=\(\left(\dfrac{5}{17}+\dfrac{12}{17}\right)+\left(\dfrac{1}{22}-\dfrac{23}{22}\right)+\dfrac{2}{3}\)

=\(\dfrac{17}{17}-\dfrac{22}{22}+\dfrac{2}{3}\)

=\(1-1+\dfrac{2}{3}\)

=0+\(\dfrac{2}{3}\)

=\(\dfrac{2}{3}\)

2/3

Ta có:\(\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{10.10}>\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\left(1\right)\)

Đặt \(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)

\(A=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\left(2\right)\)

Từ \(\left(1\right)\)và\(\left(2\right)\)suy ra

\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{10^2}>\frac{9}{22}\)

^^

\(A=47.36+64.47+15\)

\(A=47.\left(36+64\right)+15\)

\(A=47.100+15\)

\(A=4700+15\)

\(A=4715\)

\(B=27+35+65+73+75\)

\(B=\left(27+73\right)+\left(35+65\right)+75\)

\(B=100+100+75\)

\(B=275\)

\(C=37+37.15+84.37\)

\(C=37.\left(1+15+84\right)\)

\(C=37.100\)

\(C=3700\)

\(D=\frac{1}{20.21}+\frac{1}{21.22}+\frac{1}{22.23}+\frac{1}{23.24}\)

\(D=\frac{1}{20}-\frac{1}{21}+\frac{1}{21}-\frac{1}{22}+\frac{1}{22}-\frac{1}{23}+\frac{1}{23}-\frac{1}{24}\)

\(D=\frac{1}{20}-\frac{1}{24}\)

\(D=\frac{24}{480}-\frac{20}{480}\)

\(D=\frac{4}{480}=\frac{1}{120}\)

\(E=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)

\(E=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(E=1-\frac{1}{50}\)

\(E=\frac{49}{50}\)

trên đầu bài là giấu phẩy hay giấu nhân thế

\(a,2^2=4,2^3=8,2^4=16,2^5=32,2^6=64,2^7=128,2^8=256,2^9=512,2^{10}=1024\)

\(b,3^2=9,3^3=27,3^4=81,3^5=243\)

\(c,4^2=16,4^3=64,4^4=256\)

\(d,5^2=25,5^3=125,5^4=625\)

a: \(2^2=4\)

\(2^3=8\)

\(2^4=16\)

\(2^5=32\)

\(2^6=64\)

\(2^7=128\)

\(2^8=256\)

\(2^9=512\)

\(2^{10}=1024\)

b: \(3^2=9\)

\(3^3=27\)

\(3^4=81\)

\(3^5=243\)

c: \(4^2=64\)

\(4^3=256\)

\(4^4=1024\)

d: \(5^2=25\)

\(5^3=125\)

\(5^4=625\)