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Vì 13 là lẻ \(\Rightarrow\) 13, 132, 133, 134, 135, 136 là lẻ.
Mà lẻ + lẻ + lẻ + lẻ + lẻ + lẻ = chẵn nên 13 + 132 + 133 + 134 + 135 + 136 là chẵn. \(\Rightarrow\) 13 + 132 + 133 + 134 + 135 + 136 \(⋮\) 2
\(\Rightarrow\) ĐPCM
\(M=2+2^3+2^5+2^7+....+2^{51}\)
\(=\left(2+2^3\right)+\left(2^5+2^7\right)+....+\left(2^{49}+2^{51}\right)\)
\(=10+2^4\left(2+2^3\right)+....+2^{48}\left(2+2^3\right)\)
\(=10+2^4.10+...+2^{48}.10\)
\(=10\left(1+2^4+...+2^{48}\right)\Rightarrow M⋮10\)
\(=2.5.\left(1+2^4+...+2^{48}\right)\Rightarrow M⋮5\)
\(M=2+2^3+2^5+2^7+....+2^{51}.\)
\(M+2^{ }=2+2+2^3+2^5+2^7+.....+2^{51}\)
\(=\left(2+2+2^3\right)+\left(2^5+2^7+2^9\right)+....+\left(2^{47}+2^{49}+2^{51}\right)\)
\(=12+2^4\left(2+2^3+2^5\right)+......+2^{46}\left(2+2^3+2^5\right)\)
\(=12+2^4.42+....+2^{46}.42\)
\(=12+7.3.2\left(2^4+...+2^{46}\right)\)
\(\Rightarrow M=\left[12+7.3.2\left(2^4+.....+2^{46}\right)\right]-2\)
\(=10+7.3.2\left(2^4+....+2^{46}\right)\)
Ta có: \(7.3.2\left(2^4+...+2^{46}\right)⋮7\)mà 10 không chia hết cho 7
Suy M không chia hết cho 7
Cho \(A=2+2^2+2^3+2^4+...+2^{60}\)
Chứng tỏ
a, A chia hết cho 3
b, A chia hết cho 5
c, A chia hết cho 7
a) \(A=2+2^2+2^3+2^4+...+2^{60}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
\(=2\left(2+1\right)+2^3\left(2+1\right)+...+2^{59}\left(2+1\right)\)
\(=3\left(2+2^3+...+2^{59}\right)⋮3\)
Vậy \(A⋮3\)
b) \(A=2+2^2+2^3+2^4+...+2^{60}\)
\(=\left(2+2^3\right)+\left(2^2+2^4\right)+...+\left(2^{58}+2^{60}\right)\)
\(=2\left(1+2^2\right)+2^2\left(1+2^2\right)+...+2^{58}\left(1+2^2\right)\)
\(=5\left(2+2^2+...+2^{58}\right)⋮5\)
Vậy \(A⋮5\)
c) \(A=2+2^2+2^3+2^4+...+2^{60}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+..+2^{58}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+...+2^{58}\right)⋮7\)
Vậy \(A⋮7\)
Cho \(A=2+2^2+2^3+2^4+...+2^{60}\)
Chứng tỏ
a, A chia hết cho 3
b, A chia hết cho 5
c, A chia hết cho 7
a) \(A=2+2^2+2^3+2^4+...+2^{60}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{59}\right)⋮3\)
Vậy \(A⋮3\)
b1:
B=3+3^2+...+3^60=(3+3^2+3^3)+...+(3^58+3^59+3^60)=3(1+3+3^2)+...+3^58(1+3+3^2)=3*13+...+3^58*13=13(3+...+3^58) (CHIA HẾT CHO 13)
A=5+5^2+...+5^10=(5+5^2)+(5^3+5^4)+...+(5^9+5^10)=5(1+5)+...+5^9(1+5)=5*6+...+5^9*6=(5+...+5^9)*6(CHIA HẾT CHO 6)
B2: bạn kéo xuống dưới nãy mk thấy có ng làm r
b3: (2x+1)(y-5)=168
Ta có bảng sau:
| 2x+1 | 1 | 2 | 4 | 7 | 8 | 12 | 14 | 21 | 24 | 42 | 84 | 168 |
| 2x | 0 | 1 | 3 | 6 | 7 | 11 | 13 | 20 | 23 | 41 | 83 | 167 |
| x | 0 | 3 | 10 | |||||||||
| y-5 | 168 | 24 | 8 | |||||||||
| y | 173 | 29 | 13 |
(mấy ô mk để trống là loại vì x,y là số tự nhiên)
a) \(A=2+2^2+2^3+2^4+....+2^{60}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{59}+2^{60}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+....+2^{59}\left(1+2\right)\)
\(=\left(1+2\right)\left(2+2^3+...+2^{59}\right)\)
\(=3\left(2+2^3+...+2^{59}\right)\)\(⋮\)\(3\)
b) mk chỉnh lại đề
\(7^6+7^5+7^4=7^4\left(7^2+7+1\right)=7^2.57\)\(⋮\)\(57\)