Ta có: \(\frac{2n^3+n^2+7n+1}{2n-1}=\frac{\left(2n-1\right)\left(n^2+n+4\right)+5}{2n-1}=n^2+n+4+\frac{5}{2n-1}\)

Để 2n³ + n² + 7n + 1 chia hết cho 2n - 1 thì \(\frac{5}{2n-1}\in\Rightarrow\Leftarrow5⋮2n-1\Rightarrow2n-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Ta lập bảng giá trị sau:

Vậy \(n\in\left\{1;0;3;-2\right\}\)thì 2n³ + n² + 7n + 1 chia hết cho 2n - 1

\(2n^3+n^2+7n+1\)

\(=\left(2n-1\right)\left(n^2+n+4\right)+5\)

\(\Rightarrow\frac{2n^3+n^2+7n+1}{2n-1}=n^2+n+4+\frac{5}{2n-1}\)

Để vế trái nguyên thì \(2n-1\)là Ư(5).

\(\Rightarrow n=-2,0,1,3\)

a) \(n=\left\{0;1;2;3\right\}\)

b) \(n=\left\{4;5;6\right\}\)

c) \(n=6\)

d) \(n=\left\{3;4;\right\}\)

a)n thuoc{1;2;3}

b)n thuoc{1;2;3;4;5;6}

c) n thuoc{1;2;4;5;6}

d)n thuoc{1;2;3;4;5;6}

\(a)P=\left\{343;2401\right\}\)

\(b)x-\left\{x-\left[x-\left(x+1\right)\right]\right\}=x-\left\{x-\left[x-x-1\right]\right\}\)

\(=x-\left\{x-x+x+1\right\}\)

\(=x-x+x-x-1=-1\)

a) \(n=\left\{3,4\right\}\)

b)\(x-\left\{x-\left[x-\left(x-1\right)\right]\right\}\)

\(=x-\left[x-\left(x-x+1\right)\right]\)

\(=x-\left(x-1\right)\)

\(=x-x-1=-1\)

_Tần vũ_

a.1

b.1

c.1

Giải thế ai hiểu nổi hả trời???

a) n = 6

b) n = 3 ; n = 4

a)n=6

b)n=3, n=4

a: 50<2^n<100

=>32<2^n<128

=>2^n=64

=>n=6

b: 50<7^n<2500

mà \(7^n\in\left\{7;49;343;2401;...\right\}\)

nên \(7^n\in\left\{343;2401\right\}\)

=>n=3 hoặc n=4

a: \(\Leftrightarrow n+8-11⋮n+8\)

\(\Leftrightarrow n+8\in\left\{1;-1;11;-11\right\}\)

hay \(n\in\left\{-7;-9;3;-19\right\}\)

b: Đề thiếu rồi bạn

a, \(\dfrac{n-3}{n+8}=\dfrac{n+8-11}{n+8}=1-\dfrac{11}{n+8}\)

\(\Rightarrow n+8\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)

b, bạn bổ sung đề nhé