A=31 +32+33+.........38 chia hết cho 12
76+75-74 chia hết cho 11
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Bài 1:
\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)
\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)
Bài 2:
\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)
\(a,=7^4\left(7^2+7-1\right)=7^4\cdot55=7^4\cdot5\cdot11⋮11\)
a) \(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4\left(49+7-1\right)=7^4.55⋮55\)
b) \(16^5+2^{15}=\left(2^4\right)^5+2^{15}=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{15}\left(32+1\right)=2^{15}.33⋮33\)
c) \(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}=3^{28}-3^{27}-3^{26}=3^{26}\left(3^2-3-1\right)=3^{26}.5=3^{22}.3^4.5=3^{22}.405⋮405\)
a: \(=7^4\left(7^2+7-1\right)=7^4\cdot55⋮55\)
b: \(=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{15}\cdot33⋮33\)
c: \(=3^{28}-3^{27}-3^{26}=3^{26}\left(3^2-3-1\right)=3^{26}\cdot5=3^{22}\cdot405⋮405\)
4/ Chứng minh rằng :a. 76 +75 – 74 chia hết cho 11 . bạn nào giúp mình với (giải thích cho mình hiểu luôn nha các bạ... - Hoc24
\(7^6+7^5-7^4\)
\(=7^4\left(7^2+7-1\right)\)
\(=7^4\cdot55⋮11\)
a: \(B=3^1+3^2+...+3^{2010}\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(=4\left(3+3^3+...+3^{2009}\right)⋮4\)
\(B=3\left(1+3+3^2\right)+...+3^{2008}\left(1+3+3^2\right)\)
\(=13\left(3+...+3^{2008}\right)⋮13\)
b: \(C=5^1+5^2+...+5^{2010}\)
\(=5\left(1+5\right)+...+5^{2009}\left(1+5\right)\)
\(=6\left(5+...+5^{2009}\right)⋮6\)
\(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)\)
\(=31\left(5+...+5^{2008}\right)⋮31\)
c: \(D=7\left(1+7\right)+...+7^{2009}\left(1+7\right)\)
\(=8\left(7+...+7^{2009}\right)⋮8\)
\(D=7\left(1+7+7^2\right)+...+7^{2008}\left(1+7+7^2\right)\)
\(=57\left(7+...+7^{2008}\right)⋮57\)
a) 1.2.3.4.5.6...........10 + 324
= 6 ( a) 1.2.3.4.5.7...........10 + 54) chia hết cho 6
=> a) 1.2.3.4.5.6...........10 + 324 chia hết cho 6
b) 19.220 +76 = 19.2.110+2 . 38 = 38( 110+2) chia hết cho 38
=> ) 19.220 +76 chia hết cho 38
c) 15 . 3 . 999 + 49 = 45.999 + 45 + 4 = 45 ( 999 +1) +4 = 45 . 1000 + 4 chia 45 dư 4
=> 15 . 3 . 999 + 49 ko chia hết cho 45
\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(=4.\left(3+3^3+...+3^{2009}\right)\)
⇒ \(B\) ⋮ 4
b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)
\(A=3+3^2+3^3+...+3^{2012}\\ A=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\\ A=120+...+3^{2008}.120\\ A=120.\left(1+...+3^{2008}\right)⋮120\)
\(A=3^1+3^2+3^3+...+3^8\)
\(=\left(3^1+3^2\right)+\left(3^3+3^4\right)+...+\left(3^7+3^8\right)\)
\(=1\left(3^1+3^2\right)+3^2\left(3^1+3^2\right)+...+3^6\left(3^1+3^2\right)\)
\(=1.12+3^2.12+...+3^6.12\)
\(=12.\left(1+3^2+...+3^6\right)⋮12\)
Ta có: \(7^6+7^5-7^4\)
\(=7^4\left(7^2+7-1\right)\)
\(=7^4.55\)
\(=7^4.5.11⋮11\)
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