a) Ta có: \(34^{2005}-34^{2004}\)

\(=17^{2005}\cdot2^{2005}-17^{2004}\cdot2^{2004}⋮17\)

b) Ta có: \(43^{2004}+43^{2005}\)

\(=43^{2004}\left(1+43\right)\)

\(=43^{2004}\cdot44⋮11\)

c) Ta có: \(27^3+9^5=3^9+3^{10}=3^9\left(1+3\right)=3^9\cdot4⋮4\)

Câu d nữa bạn

Cho 16a + 17 b chia hết cho 11 

Mà ( 16a + 17b ) + ( 17a +16b ) = 33a + 33b = 11(3a + 3b ) chia hết cho 11

=> 17a + 16 b chia hết cho 11

2. Câu hỏi của lekhanhhung - Toán lớp 7 - Học toán với OnlineMath

a: \(G=8^8+2^{20}\)

\(=2^{24}+2^{20}\)

\(=2^{20}\left(2^4+1\right)=2^{20}\cdot17⋮17\)

b: Sửa đề: \(H=2+2^2+2^3+...+2^{60}\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{59}\left(1+2\right)\)

\(=3\left(2+2^3+...+2^{59}\right)⋮3\)

\(H=2+2^2+2^3+...+2^{60}\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)

\(=7\left(2+2^4+...+2^{58}\right)⋮7\)

\(H=2+2^2+2^3+...+2^{60}\)

\(=\left(2+2^2+2^3+2^4\right)+...+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2\left(1+2+2^2+2^3\right)+...+2^{57}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+...+2^{57}\right)⋮15\)

c: \(E=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{1989}\left(1+3+3^2\right)\)

\(=13\left(1+3^3+...+3^{1989}\right)⋮13\)

\(E=1+3+3^2+3^3+...+3^{1991}\)

\(=\left(1+3+3^2+3^3+3^4+3^5\right)+\left(3^6+3^7+3^8+3^9+3^{10}+3^{11}\right)+...+3^{1986}+3^{1987}+3^{1988}+3^{1989}+3^{1990}+3^{1991}\)

\(=364\left(1+3^6+...+3^{1986}\right)⋮14\)

\(A=3+3^2+3^3+...+3^{30}\left(1\right)\)

\(\Rightarrow A=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{28}\left(1+3\right)\)

\(\Rightarrow A=4.3+4.3^3+...+3^{28}.4=4.\left(3+3^3+...+3^{28}\right)⋮4\Rightarrow dpcm\)

\(\Rightarrow A=4.\left(3+3^3+...+3^{28}\right)=4.3.\left(1+3^2+...+3^{27}\right)=12.\left(1+3^2+...+3^{27}\right)⋮12\)

\(\Rightarrow dpcm\)

Phần Chia cho 13, bạn xem lại đề

a) Ta có a + 5b = a - b + 6b 

Vì a - b \(⋮\)6 ; 6b  \(⋮\)6 

=> a - b + 6b  \(⋮\)6

=> a + 5b  \(⋮\)6

b) Ta có a + 17b = a - b + 18b

Vì a - b \(⋮\)6 ; 18b  \(⋮\)6

=> a - b + 18b  \(⋮\)6

=> a + 17b  \(⋮\)6

c) Ta có a - 13b = a - b - 12b

Vì a - b  \(⋮\)6

12b  \(⋮\)6

=> a - b - 12b  \(⋮\)6

=> a - 13b  \(⋮\)6