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e=20052005 +1001 chia hết cho 9
vì 2005:9 dư 2=>20052005 :9 dư 2 (1)
vì 1001:9 dư 7 (2)
từ (1),(2)=>(20052005+1001):9 dư 0
=>E=20052005+1001 chia hết cho 9
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Bài 1:
\(A=\frac{24.47-22}{24+47.23}.\frac{5+\frac{5}{7}+\frac{5}{11}-\frac{5}{13}+\frac{5}{1001}}{6+\frac{6}{7}+\frac{6}{11}-\frac{6}{13}+\frac{6}{1001}}\)\(=\frac{47.23+47-22}{24.47.23}.\frac{5\left(1+\frac{1}{7}+\frac{1}{11.}-\frac{1}{13}+\frac{1}{1001}\right)}{6\left(1+\frac{1}{7}+\frac{1}{11}-\frac{1}{13}+\frac{1}{1001}\right)}\)
\(=\frac{47.23+24}{24+47.23}.\frac{5}{6}\)
\(=1.\frac{5}{6}=\frac{5}{6}\)
Bài 2:
\(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)
\(=3^{28}-3^{27}-3^{26}\)
\(=3^{22}\left(3^6-3^5-3^4\right)\)
\(=3^{22}.405\) chia hết cho 405
=>đpcm
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S=(1+3+3^2)+(3^3+3^4+3^5)+...+(3^999+3^1000+3^1001)
S=1x(1+3+9)+3^3x(1+3+9)+...+3^999x(1+3+9)
S=1x13+3^3x13+...+3^999x13
S=13x(1+3^3+...+3^999)
Vậy S chia hết cho 13
S=(1+3+3^2)+(3^3+3^4+3^5)+...+(3^999+3^1000+3^1001)
S=1x(1+3+9)+3^3x(1+3+9)+...+3^999x(1+3+9)
S=1x13+3^3x13+...+3^999x13
S=13x(1+3^3+...+3^999)
Vậy S chia hết cho 13
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a: \(6^{1001}+1=\left(6+1\right)\cdot A=7\cdot A⋮7\)
b: \(4^{2n+1}+3^{n+2}=16^n\cdot4+3^n\cdot9\)