a: Ta có: \(A=2+2^2+2^3+2^4+...+2^{100}\)

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

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

b: Ta có: \(B=4+4^2+4^3+...+4^{2022}\)

\(=4\left(1+4\right)+4^3\left(1+4\right)+...+4^{2021}\left(1+4\right)\)

\(=5\cdot\left(4+4^3+...+4^{2021}\right)⋮5\)

Dạ em cảm ơn rất nhiều

\(B=2+2^2+2^3+...+2^{100}\)

\(\Rightarrow2B=2^2+2^3+2^4+...+2^{101}\)

\(\Rightarrow2B-B=2^2+2^3+2^4+...+2^{101}-2-2^2-2^3-...-2^{100}\)

\(\Rightarrow2B-B=2^{101}-2\)

bài 1

2¹⁰¹ - 2

\(A=47.36+64.47+15\)

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

\(A=47.100+15\)

\(A=4700+15\)

\(A=4715\)

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

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

\(B=100+100+75\)

\(B=275\)

\(C=37+37.15+84.37\)

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

\(C=37.100\)

\(C=3700\)

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

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

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

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

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

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

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

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

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

A = 47 x 36 + 64 x 47 + 15

A= 47 x ( 64 + 36 ) + 15 = 47 x 100 + 15 = 4700 + 15 = 4715

vậy A= 4715

B= 27+35 + 65 + 73+ 75

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

B= 100 +100 +75 = 275

vậy B= 275

C= 37 +37 x 15 +37 x 84 

C= 37 x ( 1+15 +84 )= 37 x 100 = 3700

 vậy C= 3700

D = 1/20x21  +  1/21x22    +    1/22x23    +    1/23x24

D= 1/20   -   1/21   +    1/21  -  1/22   + 1/22   -   1/23  +   1/23   -    1/24

D= 1/20 -1/24 = 1/120 vậy D= 1/120

E= 1/1x2   +  1/2x3 + ...... + 1/49x50

E= 1/1  -   1/2    +    1/2  -   1/3  +...... + 1/49   -   1/50

E = 1 - 1/50 = 49/50 

vậy E= 49/50

 CHÚC HOK TOT

a: Tổng các số hạng là:

\(\dfrac{\left(220+1\right)\cdot220}{2}=24310\)

Ta có: A+1=2x

\(\Leftrightarrow2x=24311\)

hay \(x=\dfrac{24311}{2}\)

Bài 1 :

a) \(a.b+b.19=713\) \(\left(a;b\inℕ^∗\right)\)

\(\Rightarrow b.\left(a+19\right)=713\)

\(\Rightarrow\left(a+19\right);b\in\left\{1;23;31;713\right\}\)

\(\Rightarrow\left(a;b\right)\in\left\{\left(-18;713\right);\left(4;31\right);\left(12;23\right);\left(694;1\right)\right\}\)

\(\Rightarrow\left(a;b\right)\in\left\{\left(4;31\right);\left(12;23\right);\left(694;1\right)\right\}\left(a;b\inℕ^∗\right)\)

b) \(a.b-10.b=650\)

\(\Rightarrow b.\left(a-10\right)=650\)

\(\Rightarrow\left(a-10\right);b\in\left\{1;5;10;13;25;26;50;65;130;325;650\right\}\)

Bạn lập bảng sẽ tìm ra (a;b)...

Bài 2 :

a) \(3^4+3^5+3^6+3^7=3^4\left(1+3+3^2+3^3\right)=3^4.40\)

b) \(B=1+3+3^2+3^3+...+3^{99}\)

\(\Rightarrow B=\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)...+3^{96}.\left(1+3+3^2+3^3\right)\)

\(\Rightarrow B=40+3^4.40...+3^{96}.40\)

\(\Rightarrow B=40\left(1+3^4...+3^{96}\right)⋮40\)

\(\Rightarrow dpcm\)

a) (26-6)*(-4)+31*(-7-13)

=-80+-620

=-700

hacker 2k6

a: \(2^3-5^3:5^2+12\cdot2^2\)

\(=8-5+48\)

\(=51\)

b: \(5\cdot\left[\left(85-35:7\right):8+90\right]-5\)

\(=5\cdot\left[10+90\right]-5\)

=495

a: 23−53:52+12⋅2223−53:52+12⋅22

=8−5+48=8−5+48

=51=51

b: 5⋅[(85−35:7):8+90]−55⋅[(85−35:7):8+90]−5

=5⋅[10+90]−5=5⋅[10+90]−5

=495

a: \(280-\left(x-140\right):35=270\)

=>(x-140):35=280-270=10

=>x-140=350

=>x=350+140

=>x=490

b: \(\left(190-2x\right):35-32=16\)

=>\(\left(190-2x\right):35=32+16=48\)

=>\(190-2x=35\cdot48=1680\)

=>2x=190-1680=-1490

=>x=-745

c: \(720:\left\lbrack41-\left(2x-5\right)\right\rbrack=2^3\cdot5\)

=>\(720:\left\lbrack41-\left(2x-5\right)\right\rbrack=8\cdot5=40\)

=>41-(2x-5)=720:40=18

=>2x-5=41-18=23

=>2x=28

=>x=14

d: \(\left(x:23+45\right)\cdot37-22=2^4\cdot105\)

=>\(\left(\frac{x}{23}+45\right)\cdot37=16\cdot105+22=1702\)

=>\(\frac{x}{23}+45=46\)

=>\(\frac{x}{23}=1\)

=>x=23

e: \(\left(3x-4\right)\left(x-1\right)^3=0\)

=>\(\left[\begin{array}{l}3x-4=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac43\\ x=1\end{array}\right.\)

f: \(2^{2x-1}:4=8^3\)

=>\(2^{2x-1-2}=2^9\)

=>2x-3=9

=>2x=12

=>x=6

g: \(x^{17}=x\)

=>\(x^{17}-x=0\)

=>\(x\left(x^{16}-1\right)=0\)

=>\(\left[\begin{array}{l}x=0\\ x^{16}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^{16}=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\\ x=-1\end{array}\right.\)

h: \(\left(x-5\right)^4=\left(x-5\right)^6\)

=>\(\left(x-5\right)^6-\left(x-5\right)^4=0\)

=>\(\left(x-5\right)^4\cdot\left\lbrack\left(x-5\right)^2-1\right\rbrack=0\)

=>\(\left(x-5\right)^4\cdot\left(x-4\right)\left(x-6\right)=0\)

=>x∈{4;5;6}

i: \(\left(x+2\right)^5=2^{10}\)

=>\(\left(x+2\right)^5=\left(2^2\right)^5=4^5\)

=>x+2=4

=>x=2

k: 1+2+3+...+x=78

=>\(\frac{x\left(x+1\right)}{2}=78\)

=>x(x+1)=156

=>\(x^2+x-156=0\)

=>(x+13)(x-12)=0

=>x=-13(loại) hoặc x=12(nhận)

l: \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)

=>\(3x-16=2\cdot\frac{7^4}{7^3}=2\cdot7=14\)

=>3x=16+14=30

=>\(x=\frac{30}{3}=10\)

n: \(5^{x}:5^2=125\)

=>\(5^{x-2}=5^3\)

=>x-2=3

=>x=5

m: \(\left(x+1\right)^2=\left(x+1\right)^0\)

=>\(\left(x+1\right)^2=1\)

=>\(\left[\begin{array}{l}x+1=1\\ x+1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-2\end{array}\right.\)

o: Số số hạng của dãy số 2;4;..;52 là:

(52-2):2+1=50:2+1=25+1=26(số)

Tổng của dãy số 2;4;...;52 là:

\(\left(52+2\right)\cdot\frac{26}{2}=54\cdot13=702\)

(2+x)+(4+x)+...+(52+x)=780

=>26x+702=780

=>26x=78

=>x=3

p: \(70=2\cdot5\cdot7;80=2^4\cdot5\)

=>ƯCLN(70;80)=\(2\cdot5=10\)

70⋮x; 80⋮x

=>x∈ƯC(70;80)

=>x∈Ư(10)

mà x>8

nên x=10

q: \(12=2^2\cdot3;25=5^2;30=2\cdot3\cdot5\)

=>BCNN(12;25;30)=\(2^2\cdot3\cdot5^2=300\)

x⋮12; x⋮25; x⋮30

=>x∈BC(12;25;30)

=>x∈B(300)

mà 0<x<500

nên x=300

\(=\frac{1}{2}\left(1-\frac{1}{399}\right)=\frac{1}{2}.\frac{1}{399}=\frac{1}{798}.\)