1) Số số hạng là n 

Tổng bằng : \(\frac{n\left(n+1\right)}{2}=378\\ \Rightarrow n\left(n+1\right)=756\\ \Rightarrow n\left(n+1\right)=27.28\\ \Rightarrow n=27\)

2) a) \(n+2⋮n-1\\ \Rightarrow n-1+3⋮n-1\\ \Rightarrow3⋮n-1\)

b) \(2n+7⋮n+1\\ \Rightarrow2\left(n+1\right)+5⋮n+1\\ \Rightarrow5⋮n+1\)

c) \(2n+1⋮6-n\\ \Rightarrow2\left(6-n\right)+13⋮6-n\\ \Rightarrow13⋮6-n\)

d) \(4n+3⋮2n+6\\ \Rightarrow2\left(2n+6\right)-9⋮2n+6\\ \Rightarrow9⋮2n+6\)

a) \(25⋮n+2\left(n\in Z\right)\)

\(\Rightarrow n+2\in\left\{-1;1;-5;5;-25;25\right\}\)

\(\Rightarrow n\in\left\{-3;-1;-7;3;-27;23\right\}\)

b) \(2n+4⋮n-1\)

\(\Rightarrow2n+4-2\left(n-1\right)⋮n-1\)

\(\Rightarrow2n+4-2n+2⋮n-1\)

\(\Rightarrow6⋮n-1\)

\(\Rightarrow n-1\in\left\{-1;1;-2;2;-3;3;-6;6\right\}\)

\(\Rightarrow n\in\left\{0;2;-1;3;-2;4;-5;7\right\}\)

c) \(1-4n⋮n+3\)

\(\Rightarrow1-4n+4\left(n+3\right)⋮n+3\)

\(\Rightarrow1-4n+4n+12⋮n+3\)

\(\Rightarrow13⋮n+3\)

\(\Rightarrow n+3\in\left\{-1;1;-13;13\right\}\)

\(\Rightarrow n\in\left\{-4;-2;-15;10\right\}\)

a) n ϵ{−3;−1;−7;3;−27;23}

b) n ∈{0;2;−1;3;−2;4;−5;7}

c) n ϵ {−4;−2;−15;10}

a) Ta có: $(3n+2,5n+3)=(3n+2,2n+1)=(n+1,2n+1)=(n+1,n)=1$.

Các câu sau chứng minh tương tự.

a: ĐKXĐ: n<>1

Để \(\frac{2n-1}{n-1}\) là số nguyên thì 2n-1⋮n-1

=>2n-2+1⋮n-1

=>1⋮n-1

=>n-1∈{1;-1}

=>n∈{2;0}

b: ĐKXĐ: n<>-1

Để \(\frac{3n+5}{n+1}\) là số nguyên thì 3n+5⋮n+1

=>3n+3+2⋮n+1

=>2⋮n+1

=>n+1∈{1;-1;2;-2}

=>n∈{0;-2;1;-3}

c: ĐKXĐ: n<>-3

Để \(\frac{4n-2}{n+3}\) là số nguyên thì 4n-2⋮n+3

=>4n+12-14⋮n+3

=>-14⋮n+3

=>n+3∈{1;-1;2;-2;7;-7;14;-14}

=>n∈{-2;-4;-1;-5;4;-10;11;-17}

d: ĐKXĐ: n<>-4/3

Để \(\frac{6n-4}{3n+4}\) là số nguyên thì 6n-4⋮3n+4

=>6n+8-12⋮3n+4

=>-12⋮3n+4

=>3n+4∈{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12}

=>3n∈{-3;-5;-2;-6;-1;-7;0;-8;2;-10;8;-16}

=>n∈{\(-1;-\frac53;-\frac23;-2;-\frac13;-\frac73;0;-\frac83;\frac23;-\frac{10}{3};\frac83;-\frac{16}{3}\) }

mà n là số nguyên

nên n∈{-1;-2;0}

e: ĐKXĐ: n<>1/2

Để \(\frac{n+3}{2n-1}\) là số nguyên thì n+3⋮2n-1

=>2n+6⋮2n-1

=>2n-1+7⋮2n-1

=>7⋮2n-1

=>2n-1∈{1;-1;7;-7}

=>2n∈{2;0;8;-6}

=>n∈{1;0;4;-3}

f: \(\frac{6n-4}{3n-2}=\frac{2\left(3n-2\right)}{3n-2}=2\) là số nguyên với mọi n nguyên

g: ĐKXĐ: n<>1/3

Để \(\frac{2n+3}{3n-1}\) là số nguyên thì 2n+3⋮3n-1

=>6n+9⋮3n-1

=>6n-2+11⋮3n-1

=>11⋮3n-1

=>3n-1∈{1;-1;11;-11}

=>3n∈{2;0;12;-10}

=>n∈{2/3;0;4;-10/3}

mà n nguyên

nên n∈{0;4}

Vì : \(2n+1⋮2n+1\Rightarrow2\left(2n+1\right)⋮2n+1\Rightarrow4n+2⋮2n+1\)

Mà : \(4n+3⋮2n+1\)

\(\Rightarrow\left(4n+3\right)-\left(4n+2\right)⋮2n+1\)

\(\Rightarrow4n+3-4n-2⋮2n+1\)

\(\Rightarrow1⋮2n+1\Rightarrow2n+1=1\Rightarrow n=0\)

Vậy n = 0 thỏa mãn

ta có:

4n+3\(⋮\)2n+1

4n+2+1\(⋮\)`2n+1

2(2n+1)+1\(⋮\)2n+1

Vì 2(n+1)\(⋮\)2n+1 nên 1\(⋮\)2n+1

=>2n+1 là Ư(1)

Ư(1)={1;-1}

n={0;-1}

a) Vì \(n;n+1\) là 2 số tự nhiên liên tiếp \(\left(n< n+1\right)\)

\(\Rightarrow\left(n;n+1\right)=1\)

\(\Rightarrow UCLN\left(n;n+1\right)=1\)

b) \(4n+18=2\left(2n+9\right)⋮\left(1;2;2n+9\right)\left(n\inℕ\right)\)

Ta lại có :

 \(2n+9⋮2n+1\)

\(\Leftrightarrow2n+9-2n-1⋮2n+1\)

\(\Leftrightarrow8⋮2n+1\)

\(\Leftrightarrow2n+1\in\left\{1;2;4;8\right\}\)

\(\Leftrightarrow n\in\left\{0\right\}\)

\(\Rightarrow UCLN\left(2n+1;4n+18\right)=UCLN\left(1;18\right)=1\left(n=0\right)\)

\(\Rightarrow\left(2n+1;2n+9\right)=1\)

mà \(2n+1⋮\left(1;2n+1\right)\)

\(\Rightarrow UCLN\left(2n+1;4n+18\right)=1\)

a) \(n+1\inƯ\left(n^2+2n-3\right)\)

\(\Leftrightarrow n^2+2n-3⋮n+1\)

\(\Leftrightarrow n\left(n+1\right)+n-3⋮n+1\)

Vì \(n\left(n+1\right)⋮n+1\Rightarrow n-3⋮n+1\)

\(\Leftrightarrow n+1-4⋮n+1\)

Vì \(n+1⋮n+1\Rightarrow-4⋮n+1\Rightarrow n+1\inƯ\left(-4\right)=\left\{-1;1;-2;2;-4;4\right\}\)

Ta có bảng sau:

Vậy...

b) \(n^2+2\in B\left(n^2+1\right)\)

\(\Leftrightarrow n^2+2⋮n^2+1\)

\(\Leftrightarrow n^2+1+1⋮n^2+1\)

Vì \(n^2+1⋮n^2+1\) nên \(1⋮n^2+1\Rightarrow n^2+1\inƯ\left(1\right)=\left\{-1;1\right\}\)

Ta có bảng sau:

Vậy \(n=0\)

c) \(2n+3\in B\left(n+1\right)\)

\(\Leftrightarrow2n+3⋮n+1\)

\(\Leftrightarrow2n+2+1⋮n+1\)

\(\Leftrightarrow2\left(n+1\right)+1⋮n+1\)

Vì \(2\left(n+1\right)⋮n+1\) nên \(1⋮n+1\Rightarrow n+1\inƯ\left(1\right)=\left\{-1;1\right\}\)

Ta có bảng sau:

Vậy...

a)

Vì

Vì

Ta có bảng sau:

Vậy...

b)

Vì nên

Ta có bảng sau:

Vậy

c)

Vì nên

Ta có bảng sau:

Chọn C