a: \(\Leftrightarrow2n-1\in\left\{1;-1;3;-3\right\}\)

hay \(n\in\left\{1;0;2;-1\right\}\)

c: \(\Leftrightarrow n+1\in\left\{1;-1\right\}\)

hay \(n\in\left\{0;-2\right\}\)

a: =>\(n+2\in\left\{1;-1;7;-7\right\}\)

=>\(n\in\left\{-1;-3;5;-9\right\}\)

b: =>n-3+4 chia hết cho n-3

=>\(n-3\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(n\in\left\{4;2;5;1;7;-1\right\}\)

c: =>3n^3+n^2+9n^2-1-4 chia hết cho 3n+1

=>\(3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(n\in\left\{0;-\dfrac{2}{3};\dfrac{1}{3};-1;1;-\dfrac{5}{3}\right\}\)

d: =>10n^2-10n+11n-11+1 chia hết cho n-1

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

=>\(n\in\left\{2;0\right\}\)

a: \(d=UCLN\left(n+1;n+2\right)\)

\(\Leftrightarrow n+2-n-1⋮d\)

hay d=1

b: \(d=UCLN\left(2n+2;2n+3\right)\)

\(\Leftrightarrow2n+3-2n-2⋮d\)

hay d=1

k hộ mik nhé

TL

k hộ mik

Hoktot~

a: \(d=UCLN\left(n+1;n+2\right)\)

\(\Leftrightarrow n+2-n-1⋮d\)

hay d=1

b: \(d=UCLN\left(2n+2;2n+3\right)\)

\(\Leftrightarrow2n+3-2n-2⋮d\)

hay d=1

1)

gọi ƯC(3n-2,4n-3) là d

=>\(\hept{\begin{cases}3n-2⋮d\\4n-3⋮d\end{cases}}\Rightarrow\hept{\begin{cases}12n-8⋮d\\12n-9⋮d\end{cases}}\Rightarrow\left(12n-8\right)-\left(12n-9\right)⋮d\Rightarrow1⋮d\Rightarrow d=1;-1\)

=>ƯC(3n-2,4n-3)={1;-1}

=>\(\frac{3n-2}{4n-3}\)là p/số tối giản

vậy...

a, 4n + 23 ⋮ 2n + 3

    4n + 6 + 17  ⋮ 2n + 3

   2.(2n + 3) + 17 ⋮ 2n + 3

                       17 ⋮ 2n + 3

2n + 3 \(\in\) Ư(17) = { 1; 17}

n \(\in\) {- 1; 7}

Vì n là số tự nhiên nên n = 7

b, 3n + 11 ⋮ n  - 3

   3n - 9 + 20 ⋮ n - 3

   3.(n - 3) + 20 ⋮ n - 3

                   20 ⋮ n  -3

   n - 3 \(\in\) Ư(20) = {1; 2; 4; 5; 10; 20}

n \(\in\) {4; 5; 7; 8; 13; 23}

Ta có :

\(A=3^{4\left(n+1\right)}-4^{3\left(n+1\right)}=81^{n+1}-64^{n+1}\)

\(=\left(81-64\right)\left(81^n+81^{n-1}.64+...+81.64^{n-1}+64^n\right)\)

\(=17\left(81^n+81^{n-1}.64+...+81.64^{n-1}+64^n\right)\)chia hết cho 17

Vậy ...