\(9^{34}-27^{22}+81^{16}.\)

\(=\left(3^2\right)^{34}-\left(3^3\right)^{22}+\left(3^4\right)^{16}\)

\(=3^{68}-3^{66}+3^{64}\)

\(=3^{64}.\left(3^4-3^2+1\right)\)

\(=3^{64}.\left(81-9+1\right)\)

\(=3^{64}.73\)

\(=3^{62}.3^2.73\)

\(=3^{62}.9.73\)

\(=3^{62}.657\)

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 \(9^{34}-27^{22}+81^{16}=9^{34}-\left(3^3\right)^{22}+\left(9^2\right)^{16}\)

\(=9^{34}-3^{66}+9^{32}=9^{34}-9^{33}+9^{32}\)

\(=9^{32}\left(9^2-9+1\right)=9^{32}.73\)

\(=9^{31}.\left(8.73\right)=9^{31}.657⋮657\)

a) Sai đề. 

b) \(9^{34}-27^{22}+81^{16}\)

\(=3^{68}-3^{66}+3^{64}\)

\(=3^{64}\left(3^4-3^2+1\right)=3^{64}.73=3^{62}.9.73\)

= \(3^{62}.657⋮657\)

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 ...

SKT_NTT .Dòng thứ 3 từ trên xuống có vấn đề.

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 = 7³ x (7³ + 7² - 7) = 7⁴ x 385 = 7⁴ x 35 x 11

Vậy 7⁶ + 7⁵ - 7⁴ chia chết cho 35

b.

16⁵ + 2¹⁵ = (2⁴)⁵ + 2¹⁵ = 2²⁰ + 2¹⁵ = 2¹⁵ x (2⁵ + 1) = 2¹⁵ x 33

Vậy 16⁵ + 2¹⁵ chia hết cho 33

c.

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

Chúc bạn học tốt ^^