Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
\(A=1+2^1+2^2+...+2^{2017}\)
\(2A=2+2^2+2^3+...+2^{2018}\)
\(2A-A=2^{2018}-1hayA=2^{2018}-1\)
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
Bài 2:
\(A=3+3^2+...+3^{2018}\)
\(\Leftrightarrow3A=3^2+3^3+...+3^{2019}\)
\(\Leftrightarrow2A=3^{2019}-3\)
hay \(A=\dfrac{3^{2019}-3}{2}=\dfrac{3^{2019}+9-12}{2}=\dfrac{3\left(3^{2018}+3\right)-12}{2}\)
=>A>B
\(A=\left(6^{2019}-6^{2018}\right):6^{2018}\)
\(=6^{2019}:6^{2018}-6^{2018}:6^{2018}\)
\(=6-1\)
\(=5\)
\(B=234:\left\{3.\left[47-\left(4^2+5\right)\right]\right\}\)
\(=234:\left[3.\left(47-21\right)\right]\)
\(=234:\left(3.26\right)\)
\(=234:78\)
\(=3\)
\(D=2.\left[\left(7-3^3:3^2\right):2^2+99\right]-100\)
\(=2.\left[\left(7-3\right):4+99\right]-100\)
\(=2.\left(1+99\right)-100\)
\(=100.\left(2-1\right)\)
\(=100\)
Ta có:A=1-2+22-23-...-22017+22018
=>2A=2-22+23-24-...-22018+22019
2A+A=2-22+23-24-...-22018+22019+1-2+22-23-...-22017+22018
3A=22019+1
=>A=(22019+1):3
A= 2 + 22 + 23 + ................+ 22017 + 22018
\(\Rightarrow\)A= ( 2 + 22 ) +( 23 + 24 ) +......................+ ( 22017 + 22018 )
\(\Rightarrow\)A= 6 + 22 x (2 +22 ) + ..............22016 x ( 2 + 22 )
\(\Rightarrow\)A= 6+ 22 x 6 + .................22016 x 6
\(\Rightarrow\)A= 6 x ( 1+ 22 + ........22016 )
Vì 6 \(⋮\)3 \(\Rightarrow\)6 x ( 1+ 22 + ........22016 ) \(⋮\)3
\(\Rightarrow\)\(\Rightarrow\)A \(⋮\)3
Vậy A \(⋮\)3
Lp 7 ko bt lm toán
Lp 6
Hỳ Hỳ
..army