Ta có: \(7.5^{2n}+12.6^n\)

= \(7.5^{2n}+\left(19-7\right).6^n\)

= \(7.5^{2n}+19.6^n-7.6^n\)

= \(7\left(5^{2n}-6^n\right)+19.6^n\)

= \(7\left(25^n-6^n\right)+19.6^n\)

Có: \(19+6^n⋮19\)

\(7\left(25^n-6^n\right)⋮19\)

Vậy...................(đpcm)

Đặt \(A=7.5^{2n}+12.6^n=7.25^n+12.6^n\)

Do \(25\equiv6\left(mod19\right)\Rightarrow25^n\equiv6^n\left(mod19\right)\)

\(\Rightarrow A\equiv7.6^n+12.6^n\left(mod19\right)\)

\(\Rightarrow A\equiv19.6^n\left(mod19\right)\)

Do \(19.6^n⋮19\Rightarrow A⋮19\)

A = 7.5²ⁿ + 12.6ⁿ

A = 7.(5²)ⁿ + 12.6ⁿ

A = 7.25ⁿ + 12.6ⁿ

25  \(\equiv\) 6 (mod 19)

25ⁿ \(\equiv\) 6ⁿ (mod 19)

7    \(\equiv\) - 12 (mod 19)

⇒ 7.25ⁿ \(\equiv\) -12.6ⁿ (mod 19)

⇒ 7.25ⁿ -( -12.6ⁿ) ⋮ 19

⇒ 7.25ⁿ + 12.6ⁿ   ⋮ 19

Ta có:

\(A=7.5^{2n}+12.6^n=7.25^n+12.6^n\)

Vì \(25\equiv6\left(mod19\right)\Rightarrow25^n\equiv6^n\left(mod19\right)\)

\(\Rightarrow A\equiv7.6^n+12.6^n\left(mod19\right)\)

\(\Rightarrow A\equiv19.6^n\left(mod19\right)\)

\(\Rightarrow A\equiv0\left(mod19\right)\)

Vậy ....

1)

a)2⁵¹-1

=(2³)¹⁷-1\(⋮\)2³-1=7

Vậy 2⁵¹-1\(⋮\)7

b)2⁷⁰+3⁷⁰

=(2²)³⁵+(3²)³⁵\(⋮\)2²+3²=13

Vậy 2⁷⁰+3⁷⁰\(⋮\)13

c)17¹⁹+19¹⁷

=(BS18-1)¹⁹+(BS18+1)¹⁷

=BS18-1+BS18+1

=BS18\(⋮\)18

d)36⁶³-1\(⋮\)35\(⋮\)7

Vậy 36⁶³-1\(⋮\)7

36⁶³-1

=36⁶³+1-2

=BS37-2\(⋮̸\)37

Vậy 36⁶³-1\(⋮̸\)37

e)2⁴ⁿ-1

=(2⁴)ⁿ-1\(⋮\)2⁴-1=15

Vậy 2⁴ⁿ-1\(⋮\)15

a,  n=3k => n chia hết cho 3 => đpcm

n=3k+1 => n+5 chia hết cho 3 => đpcm

n=3k+2 => n+40 chia hết cho3 => đpcm

vậy ....

b, c tương tự

Giả sử 9a + 5b : 19
Khử a: 
3a + 8b : 19 => 9.(3a + 8b) = 27a + 72b
9a + 5b : 19 => 3.(9a + 5b) = 27a + 15b
=> (27a + 72b) - (27a + 15b) = 27a + 72b - 27a - 15b = 57b = 19.3b : 19 (1)
Khử b:
3a + 8b : 19 => 5.(3a + 8b) = 15a + 40b
9a + 5b : 19 => 8.(9a + 5b) = 72a + 40b
=> (15a + 40b) - (72a + 40b) = 15a + 40b - 72a - 40b = 57a = 19.3b : 19 (2)
Từ (1) và (2) => 9a + 5b : 19

Lời giải:

$2^3\equiv -1\pmod 9$

$\Rightarrow 2^{6n}\equiv (-1)^{2n}\equiv 1\pmod 9$

$\Rightarrow 2^{6n+2}=2^{6n}.4\equiv 4\pmod 9$

$\Rightarrow 2^{6n+2}=9k+4$ với $k$ tự nhiên.

Vì $2^{6n+2}$ chẵn nên $9k$ chẵn $\Rightarrow k$ chẵn.

Khi đó:
\(2^{2^{6n+2}}+3=2^{9k+4}+3\)

$2^9\equiv -1\pmod {19}$

$\Rightarrow 2^{9k}\equiv (-1)^k\equiv 1\pmod {19}$ (do $k$ chẵn)

$\Rightarrow 2^{9k+4}\equiv 16\pmod {19}$

$\Rightarrow 2^{2^{6n+2}}+3=2^{9k+4}+3\equiv 16+3\equiv 19\equiv 0\pmod {19}$

Vậy $2^{2^{6n+2}}+3\vdots 19$