mình đang cần gấp m.n giúp mình nha :33

chia hết cho mấy vậy bạn?

\(A=5+4^2+...+4^{2021}\\ A=4^0+4^1+...+4^{2021}\\ 4A=4^1+4^2+...+4^{2022}\\ 4A-A=\left(4^1+4^2+...+4^{2022}\right)-\left(4^0+4^1+...+4^{2021}\right)\\ 3A=4^{2022}-1\\ 3A+1=4^{2022}⋮4^{2021}\)

Lời giải:
$A-1=4+4^2+4^3+...+4^{2020}+4^{2021}$
$4(A-1)=4^2+4^3+4^4+....+4^{2021}+4^{2022}$

$\Rightarrow 4(A-1)-(A-1)=4^{2022}-4$

$3(A-1)=4^{2022}-4$

$\Rightarrow 3A+1=4^{2022}\vdots 4^{2021}$ 

\(a,\dfrac{6}{7}+\dfrac{3}{10}=\dfrac{60}{70}+\dfrac{21}{70}=\dfrac{81}{70}\\ b,\dfrac{5}{9}+\dfrac{1}{3}=\dfrac{5}{9}+\dfrac{3}{9}=\dfrac{8}{9}\\ c,\dfrac{5}{8}-\dfrac{2}{5}=\dfrac{25}{40}-\dfrac{16}{40}=\dfrac{9}{40}\\ d,\dfrac{1}{4}-\dfrac{1}{7}=\dfrac{7}{28}-\dfrac{4}{28}=\dfrac{3}{28}\)

a) Ta có A = 1 + 2¹ + 2² + ... + 2²⁰²¹

           2A = 2¹ + 2² + 2³ + ... + 2²⁰²²

Vậy 2A = 2¹ + 2² + 2³ + ... + 2²⁰²²

b) 2A - A = ( 2¹ + 2² + 2³ + ... + 2²⁰²² ) - ( 1 + 2¹ + 2² + ... + 2²⁰²¹ )

           A = 2²⁰²² - 1

Vậy A = 2²⁰²² - 1

a)

\(A=1+2^1+2^2+2^3+...+2^{2020}+2^{2021}\)

\(2A=2^1+2^2+2^3+2^4+...+2^{2021}+2^{2022}\)

b)

\(2A=2^1+2^2+2^3+...+2^{2022}\)

\(2A-A=\left(2^1+2^2+2^3+...+2^{2022}\right)-\left(1+2^1+2^2+....+2^{2021}\right)\)

\(A=2^{2022}-1\)

=> đpcm

a/

\(2A=2+2^2+2^3+...+2^{2022}\)

b/

\(A=2A-A=2^{2022}-1\)

,làm ơn giúp mik với ah

\(\left(1+\dfrac{2}{3}\right).\left(1+\dfrac{2}{4}\right).\left(1+\dfrac{2}{5}\right)....\left(1+\dfrac{2}{2020}\right).\left(1+\dfrac{2}{2021}\right)\)

= \(\dfrac{5}{3}.\dfrac{6}{4}.\dfrac{7}{5}.\dfrac{8}{6}.\dfrac{9}{7}....\dfrac{2022}{2020}.\dfrac{2023}{2021}\)

= \(\dfrac{1}{3}.\dfrac{1}{4}.2022.2023\)

= \(\dfrac{337.2023}{2}\)

= \(\dfrac{\text{681751}}{2}\)

Bài 3:

\(A=5+5^2+..+5^{12}\)

\(5A=5\cdot\left(5+5^2+..5^{12}\right)\)

\(5A=5^2+5^3+...+5^{13}\)

\(5A-A=\left(5^2+5^3+...+5^{13}\right)-\left(5+5^2+...+5^{12}\right)\)

\(4A=5^2+5^3+...+5^{13}-5-5^2-...-5^{12}\)

\(4A=5^{13}-5\)

\(A=\dfrac{5^{13}-5}{4}\)

a: \(=\dfrac{10-6}{15}\cdot\dfrac{1}{3}=\dfrac{4}{45}\)

b: \(=\dfrac{10+6}{15}\cdot\dfrac{1}{3}=\dfrac{16}{45}\)

c: \(=\dfrac{4+3}{6}\cdot\dfrac{4}{3}=\dfrac{7\cdot4}{6\cdot3}=\dfrac{28}{18}=\dfrac{14}{9}\)

d: \(=\dfrac{20-14}{35}\cdot\dfrac{9}{4}=\dfrac{6}{35}\cdot\dfrac{9}{4}=\dfrac{54}{140}=\dfrac{27}{70}\)

a (2/3 - 2/5 ) nhân 1/3

C1) 

=4/15 x 1/3

= 4/45

C2)

2/3 x 1/3 - 2/5 x 1/3

= 2/9 - 2/15

= 4/45

b (2/3 + 2/5 )nhân 1/3

GIỐNG CÂU A

c (2/3 + 1/2 ) chia 3/4

C1) =7/6 : 3/4

= 14/9

C2) 

2/3 : 3/4 + 1/2 : 3/4

= 8/9 + 2/3

= 14/9

d(4/7 - 2/5 ) chia 4/9

C1) 

= 6/35 : 4/9

= 27/70

C2)

4/7 : 4/9 - 2/5 : 4/9

= 9/7 -9/10

= 27/70

a) C1 : \(\left(\dfrac{2}{3}-\dfrac{2}{5}\right)\times\dfrac{1}{3}=\dfrac{4}{15}\times\dfrac{1}{3}=\dfrac{4}{45}\)

    C2 : \(\left(\dfrac{2}{3}-\dfrac{2}{5}\right)\times\dfrac{1}{3}=\dfrac{2}{3}\times\dfrac{1}{3}-\dfrac{2}{5}\times\dfrac{1}{3}=\dfrac{2}{9}-\dfrac{2}{15}=\dfrac{4}{45}\)

b) C1 \(\left(\dfrac{2}{3}+\dfrac{2}{5}\right)\times\dfrac{1}{3}=\dfrac{16}{15}\times\dfrac{1}{3}=\dfrac{16}{45}\)

    C2 \(\left(\dfrac{2}{3}+\dfrac{2}{5}\right)\times\dfrac{1}{3}=\dfrac{2}{3}\times\dfrac{1}{3}+\dfrac{2}{5}\times\dfrac{1}{3}=\dfrac{2}{9}+\dfrac{2}{15}=\dfrac{16}{45}\)

c) C1 \(\left(\dfrac{2}{3}+\dfrac{1}{2}\right):\dfrac{3}{4}=\left(\dfrac{2}{3}+\dfrac{1}{2}\right)\times\dfrac{4}{3}=\dfrac{7}{6}\times\dfrac{4}{3}=\dfrac{14}{9}\)

   C2 \(\left(\dfrac{2}{3}+\dfrac{1}{2}\right):\dfrac{3}{4}=\left(\dfrac{2}{3}+\dfrac{1}{2}\right)\times\dfrac{4}{3}=\dfrac{2}{3}\times\dfrac{4}{3}+\dfrac{1}{2}\times\dfrac{4}{3}=\dfrac{8}{9}+\dfrac{2}{3}=\dfrac{14}{9}\)

d) C1 \(\left(\dfrac{4}{7}-\dfrac{2}{5}\right):\dfrac{4}{9}=\left(\dfrac{4}{7}-\dfrac{2}{5}\right)\times\dfrac{9}{4}=\dfrac{6}{35}\times\dfrac{9}{4}=\dfrac{27}{70}\)

    C2 \(\left(\dfrac{4}{7}-\dfrac{2}{5}\right):\dfrac{4}{9}=\left(\dfrac{4}{7}-\dfrac{2}{5}\right)\times\dfrac{9}{4}=\dfrac{4}{7}\times\dfrac{9}{4}-\dfrac{2}{5}\times\dfrac{9}{4}=\dfrac{36}{28}-\dfrac{9}{10}=\dfrac{27}{70}\)

ta có :

\(1-\frac{2}{2.3}=\frac{2.3-2}{2.3}=\frac{1.2}{2.3}\)

tương tự : \(1-\frac{2}{3.4}=\frac{2.3}{3.4},....,1-\frac{2}{2020.2021}=\frac{2019.2020}{2020.2021}\)

Vậy \(S=\frac{1.2}{2.3}.\frac{2.3}{3.4}.....\frac{2019.2020}{2020.2021}=\frac{1.\left(2.3...2019\right)^2.2020}{2.\left(3.4....2020\right)^2.2021}=\frac{2}{2020.2021}\)