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Chứng minh rằng:\(81^7-27^9-9^{13}\) chia hết cho 45
Giải:Ta có:\(81^7-27^9-9^{13}\)
= (34)7 - (33)9 - (32)13
= 328 - 327 - 326 = 326.(32-3-1)
= 326. (9 - 3 - 1) = 326.5 = 324.32.5=324.45 chia hết cho 45
Lời giải:
a) Ta có:
\(7^6+7^5-7^4=7^{4+2}+7^{4+1}-7^4\)
\(=7^4.7^2+7^4.7-7^4=7^4(7^2+7-1)=7^4.55=11.7^4.5\vdots 11\) (đpcm)
b)
\(81^7-27^9-9^{13}=(3^4)^7-(3^3)^9-(3^2)^{13}\)
\(=3^{28}-3^{27}-3^{26}\)
\(=3^{26}(3^2-3-1)=5.3^{26}=5.3.3.3^{24}=45.3^{24}\vdots 45\) (đpcm)
a, \(7^6+7^5-7^4⋮11\)
= \(7^4.7^2+7^4.7-7^4\)
= \(7^4.\left(7^2+7-1\right)\)
= \(7^4.\left(49+7-1\right)\)
=\(7^4.55=7^4.5.11\) => chia hết cho 11
b, \(81^7\)- \(27^9\)- \(9^{13}\)
=\(\left(3^4\right)^7\)- \(\left(3^3\right)^9\) - \(\left(3^2\right)^{13}\)
= \(3^{28}-3^{27}-3^{26}\)
=\(3^{26}.\left(3^2-3-1\right)\)
=3^26.5=3^13.3^2.5=45.3^13 chia hết cho 45
a, \(10^9+10^8+10^7⋮222\)
Ta có:\(10^9+10^8+10^7=10^7.\left(10^2+10+1\right)\)
\(=10^7.111=5^7.2^7.111=5^7.2^6.2.111=5^7.2^6.222\)
Vì 222\(⋮222\Rightarrow5^7.2^6.222⋮222\)
Vậy \(10^9+10^8+10^7⋮222\)
b) 817 - 279 - 913 ⋮ 45
\(\)Ta có: \(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)
\(=3^{28}-3^{27}-3^{26}=3^{26}.\left(3^2-3-1\right)\)
\(=3^{26}.5=3^{24}.3^2.5=3^{24}.45\)
Vì \(45⋮45\Rightarrow3^{24}.45⋮45\)
Vậy \(81^7-27^9-9^{13}⋮45\)
CHÚC BẠN HỌC TỐT!!
\(a,7^6+7^5-7^4⋮55\)
\(7^4\left(7^2+7-1\right)⋮55\)
\(7^4\times55⋮55\left(dpcm\right)\)
\(8^{12}-2^{33}-2^{30}\)
\(=8^{12}-\left(2^3\right)^{11}-\left(2^3\right)^{10}\)
\(=8^{12}-8^{11}-8^{10}\)
\(=8^{10}\left(8^2-8-1\right)\)
\(=8^{10}\times55⋮55\left(dpcm\right)\)
ta có :
\(81^7-9^{13}+12^{25}+27^9-12^{24}=\left(3^4\right)^7-\left(3^2\right)^{13}+4^{25}.3^{25}+\left(3^3\right)^9-4^{24}.3^{24}\)
\(=3^{28}-3^{26}+3^{27}+4^{24}.3^{24}\left(4.3-1\right)=3^{26}\left(3^2-1+3\right)+4^{24}.3^{24}.11\)
\(=3^{26}.11+4^{24}.3^{24}.11\) mà \(\hept{\begin{cases}3^{26}.12̸\text{ không chia hết cho 16}\\4^{24}.3^{24}.11\text{ chia hết cho 16}\end{cases}}\)
Vậy biểu thức ban đầu không chia hết cho 16
a) 106 - 57
= 26 . 56 - 57
= 56 . (26 - 5)
= 56 . (64 - 5)
= 56 . 59 chia hết cho 59
=> đpcm
b) 817 - 279 - 913
= (34)7 - (33)9 - (32)13
= 328 - 327 - 326
= 326 .(32 - 3 - 1)
= 326 . (9 - 3 - 1)
= 324 . 32 . 5
= 324 . 9 . 5
= 324 . 45 chia hết cho 45
=> đpcm
c) 87 - 218
= (23)7 - 218
= 221 - 218
= 218 . (23 - 1)
= 218 (8 - 1)
= 217 . 2 . 7
= 217 . 14 chia hết cho 14
=> đpcm
d) 109 + 108 + 107
= 107 . (102 + 10 + 1)
= 57 . 27 . (100 + 10 + 1)
= 57 . 26 . 2 . 111
= 57 . 26 . 222 chia hết cho 222
=> đpcm
\(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)\(=3^{28}-3^{27}-3^{26}=3^{26}\times\left(3^2-3-1\right)=3^{26}\times5\div3^4\times5\)
Nguyễn huy hoàng ơi, bn giải thích cho mk: 326x5:34 x5 vs....
Ta có:\(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}=3^{28}-3^{27}-3^{26}\)
\(\Rightarrow3^4.3^{24}-3^3.3^{24}-3^2-3^{24}=\left(3^4-3^3-3^2\right).3^{24}=\left(81-27-9\right).3^{24}=45.3^{24}⋮45\)
Vậy \(81^7-27^9-9^{13}⋮45\)