Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(9^{34}-27^{22}+81^{16}.\)
\(=\left(3^2\right)^{34}-\left(3^3\right)^{22}+\left(3^4\right)^{16}\)
\(=3^{68}-3^{66}+3^{64}\)
\(=3^{64}.\left(3^4-3^2+1\right)\)
\(=3^{64}.\left(81-9+1\right)\)
\(=3^{64}.73\)
\(=3^{62}.3^2.73\)
\(=3^{62}.9.73\)
\(=3^{62}.657\)
Vì \(657⋮657\) nên \(3^{62}.657⋮657.\)
\(\Rightarrow9^{34}-27^{22}+81^{16}⋮657\left(đpcm\right).\)
Chúc bạn học tốt!
\( {9^{34}} - {27^{22}} + {81^{16}}\\ = {\left( {{3^2}} \right)^{34}} - {\left( {{3^3}} \right)^{22}} + {\left( {{3^4}} \right)^{16}}\\ = {3^{68}} - {3^{66}} + {3^{64}}\\ = {3^{62}}\left( {{3^6} - {3^4} + {3^2}} \right)\\ = {3^{62}}\left( {729 - 81 + 9} \right)\\ = {3^{63}}.657\)
chia hết cho $657$
Ta có :
934 - 2722 + 8116
= ( 32 )64 - ( 33 )22 + ( 34 )16
= 368 - 366 + 364
= 368 . ( 34 - 32 + 1 )
= 368 . 73
= 366 . ( 32 . 73 )
= 366 . 657 \(⋮\)657
Vậy ...
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) Sai đề.
b) \(9^{34}-27^{22}+81^{16}\)
\(=3^{68}-3^{66}+3^{64}\)
\(=3^{64}\left(3^4-3^2+1\right)=3^{64}.73=3^{62}.9.73\)
= \(3^{62}.657⋮657\)
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)
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.
165 + 215 = (24)5 + 215 = 220 + 215 = 215 x (25 + 1) = 215 x (32 + 1) = 215 x 33
Vậy 1615 + 215 chia hết cho 33
b.
817 - 279 - 913 = (34)7 - (33)9 - (32)13 = 328 - 327 - 326 = 322 x (36 - 35 - 34) = 322 x 405
Vậy 817 - 279 - 913 chia hết cho 405
Ta có \(9^{34}-27^{22}+81^{16}=9^{34}-\left(3^3\right)^{22}+\left(9^2\right)^{16}\)
\(=9^{34}-3^{66}+9^{32}=9^{34}-9^{33}+9^{32}\)
\(=9^{32}\left(9^2-9+1\right)=9^{32}.73\)
\(=9^{31}.\left(8.73\right)=9^{31}.657⋮657\)