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Bài 1:
a) \(7^6+7^5-7^4\)
\(=7^4.7^2+7^4.7-7^4\)
\(=7^4.\left(7^2+7-1\right)\)
\(=7^4.\left(49+7-1\right)\)
\(=7^4.55\)
\(=7^4.5.11\)
Vì \(11⋮11\) nên \(7^4.5.11⋮11\)
\(\Rightarrow7^6+7^5-7^4⋮11.\)
b) \(81^7-27^9-9^{13}\)
\(=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)
\(=3^{28}-3^{27}-3^{26}\)
\(=3^{26}.\left(3^2-3-1\right)\)
\(=3^{26}.5\)
\(=3^{13}.3^2.5\)
\(=3^{13}.45\)
Vì \(45⋮45\) nên \(3^{13}.45⋮45\)
\(\Rightarrow81^7-27^9-9^{13}⋮45.\)
Chúc bạn học tốt!
a) 7^6+7^5-7^4 chia hết cho 11
= 7^4 ( 7^2 + 7 - 1 )
= 7^4 ( 49 +7 - 1 )
= 7^4 + 55
= 7^4 x 5 x 11 chia hết cho 11 ( đpcm )
b ) 81^7-27^9-9^13 chia hết cho 45
= (3^4)^7 - ( 3^3)^9 - ( 3^2 )^13
= 3^ 28- 3 ^27 - 3^26
= 3 ^26 x ( 3^2 - 3^1 - 3^0 )
= 3^24 x 9 x5
= 3 ^24 x 45 chia hết cho 45 ( đpcm )
\(a,7^6+7^5-7^4⋮55\)
\(7^4\left(7^2+7-1\right)⋮55\)
\(7^4\times55⋮55\left(dpcm\right)\)
\(8^{12}-2^{33}-2^{30}\)
\(=8^{12}-\left(2^3\right)^{11}-\left(2^3\right)^{10}\)
\(=8^{12}-8^{11}-8^{10}\)
\(=8^{10}\left(8^2-8-1\right)\)
\(=8^{10}\times55⋮55\left(dpcm\right)\)
b) 817 - 279 -913 chia hết cho 405
Ta có: 817 - 279 -913 = 328- 327-326
= 326(32-3-1)
= 326. 5 = 322. 405 chia hết cho 405 (đpcm)
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
Giải:
a) Ta có:
\(7^6+7^5-7^4\)
\(=7^4\left(7^2+7-1\right)\)
\(=7^4.55⋮55\)
Vậy ...
b) Ta có:
\(16^5+2^{15}\)
\(=\left(2^4\right)^5+2^{15}\)
\(=2^{20}+2^{15}\)
\(=2^{15}\left(2^5+1\right)\)
\(=2^{15}.33⋮33\)
Vậy ...
c) \(81^7-27^9-9^{13}\)
\(=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)
\(=3^{28}-3^{27}-3^{26}\)
\(=3^{26}\left(3^2-3-1\right)\)
\(=3^{26}.5⋮5⋮405\)
Vậy ...
Chúc bạn học tốt!
a) 76 +75 -74
=74.72 +74.7-74
=74.(72+7-1)
=74.55⋮55
b) 165+215
=(24)5 +215
=220+215
=215.25+215
=215.(25+1)
=215.33⋮33
c)817-279-913
=(34)7-(33)9......(làm tương tự)
a) ta có : \(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.\left(49+7-1\right)=7^4.55⋮55\)
\(\Rightarrow7^4.55\) chia hết cho \(55\) \(\Leftrightarrow7^6+7^5-7^4\) chia hết cho \(55\)
vậy \(7^6+7^5-7^4\) chia hết cho \(55\) (đpcm)
b) ta có \(16^5+2^{15}=\left(2^4\right)^5+2^{15}=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{15}.\left(32+1\right)=2^{15}.33⋮33\)
\(\Rightarrow2^{15}.33\) chia hết cho \(33\) \(\Leftrightarrow16^5+2^{15}\) chia hết cho \(33\)
vậy \(16^5+2^{15}\) chia hết cho \(33\) (đpcm)
c) ta có \(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}=3^{28}-3^{27}-3^{26}\)
\(=3^{22}\left(3^6-3^5-3^4\right)=3^{22}\left(729-243-81\right)=3^{22}.405⋮405\)
\(\Rightarrow3^{22}.405\) chia hết cho \(405\) \(\Leftrightarrow81^7-27^9-9^{13}\) chia hết cho \(405\)
vậy \(81^7-27^9-9^{13}\) chia hết cho \(405\) (đpcm)
\(a.\)
\(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55⋮55\)
\(b.\)
\(16^5+2^{15}=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{15}.33⋮33\)
\(c.\)
Ta có : \(405=3^4.5\)
\(\Rightarrow81^7-27^9-9^{13}=3^{28}-3^{27}-3^{26}=3^{26}\left(3^2-3-1\right)=3^{26}.5⋮405\)
a ) 76 + 75 - 74
= 74 ( 72 + 7 - 1 )
= 74. 55 chia hết cho 55
b ) 165 + 215
= ( 24 ) 5 + 215
= 220 + 215
= 215 ( 25 + 1 )
= 215 . 33 chia hết cho 33
c ) 817 - 279 - 913
= ( 34 )7 - ( 33 )9 - ( 32 )13
= 328 - 327 - 326
= 326 ( 32 - 3 - 1 )
= 326 . 5
= 322 . 34 . 5
= 322 . 81 . 5
= 322 . 405 chia hết cho 405
a,Ta có \(55=11\times5\)
\(^{7^6+7^5-7^4=\left(...49\right)+\left(...07\right)-\left(...01\right)=\left(....55\right)}\)
Vậy số trên chia hết cho 55
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