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

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

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

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

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

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

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

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

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

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) \(9^{34}-27^{22}+81^{16}\)

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

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

= \(3^{62}.657⋮657\)

a) Ta có : 8¹⁵ - 2⁴³ + 2⁴¹

       = (2³)¹⁵ - 2⁴³ + 2⁴¹

       = 2^3.15 - 2⁴³ + 2⁴¹

       = 2⁴⁵ - 2⁴³ + 2⁴¹

       = 2⁴¹.(2⁴ - 2² + 1)

       = 2⁴¹. 13 \(⋮\)13

=>  8¹⁵ - 2⁴³ + 2⁴¹ \(⋮\)13 (đpcm)

b) Ta có : 9³⁴ - 27²² + 81⁶

            = (3²)³⁴ - (3³)²² + (3⁴)¹⁶

             = 3^2.34 - 3^3.22 + 3^4.16

             = 3⁶⁸ - 3⁶⁶ + 3⁶⁴

             = 3⁶⁴.(3⁴ - 3² + 1)

             = 3⁶² . 3². 73

             = 3⁶² . 9 . 73

             = 3⁶² . 657 \(⋮\)657

=> 9³⁴ - 27²² + 81⁶ \(⋮\)657 (đpcm)

\(a,16^5+2^{15}=\left(2^4\right)^5+2^{15}=2^{20}+2^{15}=2^{15}.\left(2^5+1\right)=2^{15}.33\) luôn chia hết cho 33 (đpcm)

\(b,81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}=3^{28}-3^{27}-3^{26}\)

\(=3^{26}.\left(3^2-3-1\right)=3^{26}.5=3^{22}.3^4.5=3^{22}.405\) chia hết cho 405 (đpcm)