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\(A=1+2^1+2^2+...+2^{2017}\)
\(2A=2+2^2+2^3+...+2^{2018}\)
\(2A-A=2^{2018}-1hayA=2^{2018}-1\)
2; 3 tuong tu
1) A = 1 + 2 + 22 + 23 + .... + 22018
2A = 2 + 22 + 23 + 24 + ..... + 22019
2A - A = ( 2 + 22 + 23 + 24 + ..... + 22019 ) - ( 1 + 2 + 22 + 23 + .... + 22018 )
Vậy A = 22019 - 1
2) B = 1 + 3 + 32 + 33 + ..... + 32018
3A = 3 + 32 + 33 + ...... + 32019
3A - A = ( 3 + 32 + 33 + ...... + 32019 ) - ( 1 + 3 + 32 + 33 + ..... + 32018 )
2A = 32019 - 1
Vậy A = ( 32019 - 1 ) : 2
3) C = 1 + 4 + 42 + 43 + ...... + 42018
4A = 4 + 42 + 43 + ...... + 42019
4A - A = ( 4 + 42 + 43 + ...... + 42019 ) - ( 1 + 4 + 42 + 43 + ...... + 42018 )
3A = 42019 - 1
Vậy A = ( 42019 - 1 ) : 3
\(2^{2018}=2^{2016}\cdot2^2=\left(2^4\right)^{504}\cdot4=16^{604}\cdot4=\overline{.....6}\cdot4=\overline{....4}\)
\(3^{2018}=3^{2016}\cdot3^2=\left(3^4\right)^{504}\cdot9=81^{504}\cdot9=\overline{.....1}\cdot9=\overline{....9}\)
\(7^{2019}=7^{2016}\cdot7^3=\left(7^4\right)^{504}\cdot\overline{.....7}=\overline{.....1}\cdot\overline{....7}=\overline{.....7}\)
\(8^{2021}=8^{2020}\cdot8=\left(8^4\right)^{505}\cdot8=\overline{....6}\cdot8=\overline{......8}\)
\(9^{2023}=9^{2022}\cdot9=\left(9^2\right)^{1011}\cdot9=\overline{.....1}\cdot9=\overline{.....9}\)
Bài giải
Ta có :
\(2^{2018}=2^{2016}\cdot2^2=\left(2^4\right)^{504}\cdot4=\overline{\left(...6\right)}^{504}\cdot4=\overline{\left(...6\right)}\cdot4=\overline{\left(...4\right)}\)
Vậy ...
\(3^{2018}=3^{2016}\cdot3^2=\left(3^4\right)^{504}\cdot9=\overline{\left(...1\right)}^{504}\cdot9=\overline{\left(...1\right)}\cdot9=\overline{\left(...9\right)}\)
Vậy ...
\(7^{2019}=7^{2016}\cdot7^3=\left(7^4\right)^{504}\cdot7^3=\overline{\left(...1\right)}^{504}\cdot343=\overline{\left(...1\right)}\cdot3=\overline{\left(...3\right)}\)
Vậy ...
\(8^{2021}=8^{2020}\cdot8=\left(8^4\right)^{505}\cdot8=\overline{\left(...6\right)}^{505}\cdot8=\overline{\left(...6\right)}\cdot8=\overline{\left(...8\right)}\)
Vậy ...
\(9^{2023}=9^{2022}\cdot9=\left(9^2\right)^{1011}\cdot9=\overline{\left(...1\right)}^{1011}\cdot9=\overline{\left(...1\right)}\cdot9=\overline{\left(...9\right)}\)
Vậy ...
\(b,2^3+3x=2018\)
\(\Rightarrow8+3x=2018\)
\(\Rightarrow3x=2018-8=2010\)
\(\Rightarrow x=2010:3=670\)
1.
a)8102-2102
= 82 .8100 - 22. 2100
=64.(84)25-4.(24)25
=64 . ...625 - 4 . ...625
=....4 - ...4
.=...0 chia hết cho 10
b)34n+1+2
=(34)n+1 + 2
= ....1 + 2
=....3 chia hết cho 3
2.
a)C = 2.1+2.3+...+2.32004
C = 2.(1+3+...+32004)
đặt D=1+3+..+32004
3D=3+ .....+32005
3D - D=32005 - 3
2D=32005-1
2D= (34)501.3 - 1
2D = 81501 .3 - 1
D= (...1 . 3- 1):2
D = (...3 - 1) :2
D= ...2 : 2
D=....1
b)B= 1+ 3+...+3300
3B= 3+...+3101
2D = 3101 - 1
D= (3101 - 1):2
D=(3100.3-1):2
D=[(34)25. 3 -1]:2
D= [...125.3-1]:2
D= [...3-1]:2
D=...2:2
D=....1
A=1+2+22+23+...+22018+22019
>2A=2(1+2+22+23+...+22018+22019)
=>2A=2+22+23+...+22018+22019
=>2A-A=(2+22+23+...+22019+22020)-(1 + 2 + 22 + 23 + ... + 22018 + 22019)
=>A=22020-1
B=1 + 32 + 34 + 36 +...+ 32018 + 32020
=>9B=3(1 + 32 + 34 + 36 +...+ 32018 + 32020)
=>9B=3+32 + 34 + 36 +...+ 32020 + 32022
=>9B-B=(3+32 + 34 + 36 +...+ 32018 + 32020)-(1 + 32 + 34 + 36 +...+ 32018 + 32020)
=.8B=32022-1
=>B=32022:8-1
giúp mình với ạ, mình đag cần gấp, ai đúng mình tích. giải rõ ràng nha
A=B5 + C6
A=C1
Mình chỉ giải được vậy thôi!