Lời giải:
$M=3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}=3^{n+1}.3^2+3^{n+1}+2^{n+2}.2+2^{n+2}$

$=3^{n+1}(9+1)+2^{n+2}(2+1)$

$=3^{n+1}.10+2^{n+2}.3$

$=6.3^n.5+6.2^{n+1}=6(3^n.5+2^{n+1})\vdots 6$ (đpcm)

A = (3ⁿ⁺¹ + 3ⁿ⁺³) + (2ⁿ⁺² + 2ⁿ⁺³)

= 3ⁿ⁺¹.(1 + 3²) + 2.(2ⁿ⁺¹ + 2ⁿ⁺²)

= 3ⁿ⁺¹.10 + 2.(2ⁿ⁺¹ + 2ⁿ⁺²)

= 3ⁿ⁺¹.5.2 + 2.(2ⁿ⁺¹ + 2ⁿ⁺²)

= 2.(3ⁿ⁺¹.5 + 2ⁿ⁺¹ + 2ⁿ⁺²) ⋮ 2   (1)

A = (3ⁿ⁺¹ + 3ⁿ⁺³) + (2ⁿ⁺² + 2ⁿ⁺³)

= 3.(3ⁿ + 3ⁿ⁺²) + 2ⁿ⁺².(1 + 2)

= 3.(3ⁿ + 3ⁿ⁺²) + 2ⁿ⁺².3

= 3.(3ⁿ + 3ⁿ⁺² + 2ⁿ⁺²) ⋮ 3   (2)

Từ (1) và (2) ⇒ A ⋮ 2 và A ⋮ 3

⇒ A ⋮ 6

\(A=3^{n+1}+9.3^{n+1}+2^n.4+2^n.8\)

\(=3^{n+1}.10+4.2^n.3\)

\(=3^n.6.5+2^n.2.6⋮6\)

\(\Rightarrow A⋮6\left(đpcm\right)\)

A= (3^n) * 3 + (3^n)*(3^3)+(2^n)*(2^2)+(2^n)*(2^3)

A= (3^n)*(3+9)+(2^n)*(4+8)

A= 12*[(3^n)+(2^n)]

do 12 chia hết cho 6 nên A chia hết cho 6 ( đpcm)

\(=3^3.3^n+3.3^n+2^3.2^n+2^2.2^n=\)

\(=3^n\left(3^3+3\right)+2^n\left(2^3+2^2\right)=30.3^n+12.2^n=\)

\(=6\left(5.3^n+2.2^n\right)⋮6\)

\(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)

\(=3^{n+1}\left(9+3\right)+2^{n+2}\left(8+4\right)\)

\(=12.3^{n+1}+12.2^{n+2}=12.\left(3^{n+1}+2^{n+2}\right)\)

mà 12⋮6

\(\Rightarrow12.\left(3^{n+1}+2^{n+2}\right)⋮6\Rightarrow dpcm\)

Chụp ảnh hoặc sử dụng gõ công thức nhé bạn. Để vầy khó hiểu lắm

\(a,lim\dfrac{2n^2+1}{3n^3-3n+3}\)

\(=lim\dfrac{\dfrac{2}{n}+\dfrac{1}{n^3}}{3-\dfrac{3}{n^2}+\dfrac{3}{n^3}}=0\)

\(\lim\dfrac{-3n^3+1}{2n+5}=\lim\dfrac{-3n^2+\dfrac{1}{n}}{2+\dfrac{5}{n}}=\dfrac{-\infty}{2}=-\infty\)

\(\lim\dfrac{n^3-2n+1}{-3n-4}=\lim\dfrac{n^2-2+\dfrac{1}{n}}{-3-\dfrac{4}{n}}=\dfrac{+\infty}{-3}=-\infty\)

1) 3n ⋮ 2n - 5

=> 2(3n) - 3(2n - 5)  ⋮ 2n - 5

=> 6n - 6n + 15 ⋮ 2n - 5

=> 15 ⋮ 2n - 5

=> 2n-5 ϵ Ư(15)

Ư(15) = {1;-1;3;-3;5;-5;15;-15}

=> n={3;2;4 ;1;5;0;10;-5}

nhớ nha

1) 3n ⋮ 2n - 5

=> 2(3n) - 3(2n - 5)  ⋮ 2n - 5

=> 6n - 6n + 15 ⋮ 2n - 5

=> 15 ⋮ 2n - 5

=> 2n-5 ϵ Ư(15)

Ư(15) = {1;-1;3;-3;5;-5;15;-15}

=> n={3;2;4 ;1;5;0;10;-5}

B = 3ⁿ⁺³ + 2ⁿ⁺³ + 3ⁿ⁺¹ + 2ⁿ⁺²

= (3ⁿ⁺³ + 3ⁿ⁺¹) + (2ⁿ⁺³ + 2ⁿ⁺²)

= 3ⁿ⁺¹.(3² + 1) + 2(2ⁿ⁺² + 2ⁿ⁺¹)

= 3ⁿ⁺¹.10 + 2.(2ⁿ⁺² + 2ⁿ⁺¹)

= 2.3ⁿ⁺¹.5 + 2.(2ⁿ⁺² + 2ⁿ⁺¹)

= 2.(3ⁿ⁺¹.6 + 2ⁿ⁺² + 2ⁿ⁺¹) ⋮ 2 (1)

B = (3ⁿ⁺³ + 3ⁿ⁺¹) + (2ⁿ⁺³ + 2ⁿ⁺²)

= 3.(3ⁿ⁺² + 3ⁿ) + 2ⁿ⁺².(2 + 1)

= 3.(3ⁿ⁺² + 3ⁿ) + 2ⁿ⁺².3

= 3.(3ⁿ⁺² + 3ⁿ + 2ⁿ⁺²) ⋮ 3 (2)

Từ (1) và (2) ⇒ B ⋮ 6

Mng ơi giúp mình với ạ

                                                 a

1: =>3n-12+17 chia hết cho n-4

=>\(n-4\in\left\{1;-1;17;-17\right\}\)

hay \(n\in\left\{5;3;21;-13\right\}\)

2: =>6n-2+9 chia hết cho 3n-1

=>\(3n-1\in\left\{1;-1;3;-3;9;-9\right\}\)

hay \(n\in\left\{\dfrac{2}{3};0;\dfrac{4}{3};-\dfrac{2}{3};\dfrac{10}{3};-\dfrac{8}{3}\right\}\)

4: =>2n+4-11 chia hết cho n+2

=>\(n+2\in\left\{1;-1;11;-11\right\}\)

hay \(n\in\left\{-1;-3;9;-13\right\}\)

5: =>3n-4 chia hết cho n-3

=>3n-9+5 chia hết cho n-3

=>\(n-3\in\left\{1;-1;5;-5\right\}\)

hay \(n\in\left\{4;2;8;-2\right\}\)

6: =>2n+2-7 chia hết cho n+1

=>\(n+1\in\left\{1;-1;7;-7\right\}\)

hay \(n\in\left\{0;-2;6;-8\right\}\)