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a)A=(2+22)+(23+24)+...(29+210)
A=2(2+1)+23(1+2)+....+29(2+1)
A=3(2+23+25+27+29)
Vay A chia het cho 3(khi chia 3 duoc 2+23+25+27+29du 0)
b)A=(2+22+23+24+25)+(26+27+28+29+210)
A=2(1+2+22+23+24)+26(1+2+22+23+24)
A=31(2+26) luon chia het cho 31 :))
\(A=1+3+3^2+.....+3^{11}\)
\(A=\left(1+3+3^2\right)+....+\left(3^9+3^{10}+3^{11}\right)\)
\(A=\left(3^0.1+3^0.3+3^0.3^2\right)+....+\left(3^9.1+3^9.3+3^9.3^2\right)\)
\(A=1.\left(1+3+3^2\right)+....+3^9\left(1+3+3^2\right)\)
\(A=1.13+....+3^9.13\)
\(A=13.\left(1+....+3^9\right)⋮13\left(đpcm\right)\)
1.
\(A=7+7^2+7^3+...+7^{78}\)
\(=\left(7+7^2\right)+\left(7^3+7^4\right)+...+\left(7^{77}+7^{78}\right)\)
\(=7\left(1+7\right)+7^3\left(1+7\right)+...+7^{77}\left(1+7\right)\)
\(=7\cdot8+7^3\cdot8+...+7^{77}\cdot8\)
\(=\left(7+7^3+...+7^{77}\right)\cdot8\) chia hết cho 8
Vậy A chia hết cho 8 (đpcm)
\(A=3+3^2+3^3+...+3^{155}\)
\(=\left(3+3^2+3^3+3^4+3^5\right)+...+\left(3^{151}+3^{152}+3^{153}+3^{154}+3^{155}\right)\)
\(=3\left(1+3+3^2+3^3+3^4\right)+...+3^{151}\left(1+3+3^2+3^3+3^4\right)\)
\(=\left(3+...+3^{151}\right)\cdot121\) chia hết cho 121
Vậy A chia hết cho 121 (đpcm)
a) (x-14):2=24-3
(x-14):2 = 13
x-14 = 13.2
x-14 = 26
x = 26 + 14
x = 40
b) x572 = x <=> x = 1 hoặc 0
a, b làm như trên nha, còn mấy bìa còn lại :
M=1+2+22+...+211
M = \(\left(1+2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}+2^{11}\right)\)
M = (1+2+4+8+16+32) + 26( 1 + 2 + 22+23+24+25)
M = 63 + 26.63
M = 63 ( 1+ 26)
M= 9.7 (1 + 2^6) chia hết cho 9 => M chia hết cho 9
S=3 + 32 +33 +.....+ 39
S = \(\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\left(3^7+3^8+3^9\right)\)
S = \(3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^7\left(1+3+3^2\right)\)
S= 3. 13 + 3^4.13 + 3^7.13
S= 13 ( 3 +3^4+3^4) chia hết cho 13 => S chia hết cho 13
M= 2+ 22 + 23+....+210
M= \(\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)
M = \(2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)
\(M=2.3+2^3.3+...+2^9.3\)
M = 3( 2+ 2^3 +...+ 2^9) chia heets cho 3
=> M chia hết cho 3
A= 7+ 72 + 73 +.....+78
A= \(\left(7+7^2+7^3+7^4\right)+\left(7^5+7^6+7^7+7^8\right)\)
A= \(7\left(1+7+7^2+7^3\right)+7^5\left(1+7+7^2+7^3\right)\)
A= 7. 400 + 7^5 . 400
A = 400( 7+7^5)
A = 5 . 80 ( 7+7^5) chia hết cho 5 => A chia hết cho 5
minh chi lam dc cau a thoi nha nhung hay t i c k cho minh
3 + 32 = 12 chia het cho 4 3 + 32 + 33 + .......+39 + 310 = 30 .[ 3+32 ] + 32 . [ 3 + 32 ] + ....+38 . [ 3 + 32 ]
=30 . 12 + 32 . 12 +.....+ 38 . 12 = 12.[30 + 32 +....+ 38 ]
vi 12 chia het cho 4 nen 12 nhan voi so tu nhien nao thi so do cung chia het cho 4 nen A chia het cho 4
1/a/ \(A=2+2^2+2^3+....+2^{10}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+....+\left(2^9+2^{10}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+....+2^9\left(1+2\right)\)
\(=2.3+2^3.3+....+2^9.3\)
\(=3\left(2+2^3+.....+2^9\right)⋮3\)
\(\Leftrightarrow A⋮3\left(đpcm\right)\)
b/ \(A=2+2^2+2^3+....+2^{10}\)
\(=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{10}\right)\)
\(=2\left(1+2+2^2+2^3+2^4\right)+2^6.\left(1+2+2^2+2^3+2^4\right)\)
\(=2.31+2^6.31\)
\(=31\left(2+2^6\right)⋮31\)
\(\Leftrightarrow A⋮31\left(đpcm\right)\)
2/ Với mọi n là số tự nhiên thì \(n\) có hai dạng :
\(\left[{}\begin{matrix}n=2k\\n=2k+1\end{matrix}\right.\)
+) \(n=2k\Leftrightarrow B=\left(n+4\right)\left(n+7\right)=\left(2k+4\right)\left(2k+7\right)\)
Mà \(2k+4⋮2\)
\(\Leftrightarrow\left(2k+4\right)\left(2k+7\right)⋮2\)
\(\Leftrightarrow B\) là số chẵn
+) \(n=2k+1\Leftrightarrow B=\left(n+4\right)\left(n+7\right)=\left(2k+1+4\right)\left(2k+1+7\right)=\left(2k+5\right)\left(2k+8\right)\)
Mà \(2k+8⋮2\)
\(\Leftrightarrow\left(2k+5\right)\left(2k+8\right)⋮2\)
\(\Leftrightarrow B\) là số chẵn
Vậy...
1/
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)
\(A=2.3+2^3.3+2^5.5+...+2^9.3=3.\left(2+2^3+...+2^9\right)\)
Do \(3⋮3\Rightarrow A⋮3\)
\(A=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)\)
\(A=2.31+2^6.31=31\left(2+2^6\right)\)
Do \(31⋮31\Rightarrow A⋮31\)
2/ \(B=\left(n+4\right)\left(n+7\right)\)
Nếu n chẵn, đặt \(n=2k\Rightarrow B=\left(2k+4\right)\left(2k+7\right)=2\left(k+2\right)\left(2k+7\right)\)
Do 2 chẵn nên B chẵn
Nếu n lẻ, đặt \(n=2k+1\Rightarrow B=\left(2k+5\right)\left(2k+8\right)=2\left(2k+5\right)\left(k+4\right)\)
2 chẵn nên B chẵn
Vậy B luôn chẵn với mọi n
3/ Đề là B(112) hay B(121) bạn?
Chia hết cho3
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