cho A= 1+6+6^2+....+6^9.CMR:A Chia43 du1
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a) \(81^5+27^6-9^8=\left(3^4\right)^5+\left(3^3\right)^6-\left(3^2\right)^8\)
\(=3^{20}+3^{18}-3^{16}\)
\(=3^{16}\left(3^4+3^2-1\right)\)
\(=3^{16}\left(81+9-1\right)\)
\(=3^{16}.89\)
\(\Rightarrow81^5+27^6-9^8⋮89\)
b) \(32^6+16^7+8^9=\left(2^5\right)^6+\left(2^4\right)^7+\left(2^3\right)^8\)
\(=2^{30}+2^{28}+2^{27}\)
\(=2^{27}\left(2^3+2+1\right)\)
\(=2^{27}\left(8+2+1\right)\)
\(=2^{27}.11\)
\(\Rightarrow32^6+16^7+8^9⋮11\)
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a, A = 1 + 5 +52 + .. + 511
A = ( 1+5 ) + ( 52 + 53) +...+ ( 510 + 511)
A = 6 + 52. 6 + ... + 510 .6
A = 6 . (1+52 + ...+ 510 )
=> A \(⋮\) 6
b, A = 1 + 5 +52 + .. + 511
A = ( 1 + 5 +52 ) + ( 53 + 54 +55 ) + ... + ( 59 + 510 + 511)
A= 31 + 31 . 53+ ... + 31.59
A = 31 . ( 1 + 53 + ... + 59 )
=> A\(⋮\) 31
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Vì
\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{99}{100}< \frac{100}{101}\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
\(\Rightarrow a< b\)
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Mấy bạn làm hộ mình nha , bài khó quá không biết làm thế nào nữa.Xin trân thành cảm ơn nếu các bạn làm chi tiết.
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a: \(=3^{28}-3^{27}-3^{26}\)
\(=3^{26}\left(3^2-3-1\right)=3^{26}\cdot5=3^{24}\cdot45⋮45\)
b: \(=3^n\cdot27+3^n\cdot3+2^n\cdot8+2^n\cdot4\)
\(=3^n\cdot5\cdot6+2^n\cdot2\cdot6\)
\(=6\left(3^n\cdot5+2^{n+1}\right)⋮6\)
\(A=1+6+6^2+...+6^9\)
\(=1+\left(6+6^2+6^3\right)+\left(6^4+6^5+6^6\right)+\left(6^7+6^8+6^9\right)\)
\(=1+6\left(1+6+6^2\right)+6^4\left(1+6+6^2\right)+6^7\left(1+6+6^2\right)\)
\(=1+\left(1+6+6^2\right)\left(6+6^4+6^7\right)\)
\(=1+43\left(6+6^4+6^7\right)\)
Ta thấy \(43\left(6+6^4+6^7\right)⋮43\)
nên A chia 43 dư 1