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) 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 \(2n+1⋮\left(1;2n+1\right)\)

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

Chọn C

a,

Gọi \(d=ƯC\left(n+1;2n+3\right)\) với \(d\in N\)

\(\Rightarrow\left\{{}\begin{matrix}n+1⋮d\\2n+3⋮d\end{matrix}\right.\) \(\Rightarrow2n+3-2\left(n+1\right)⋮d\)

\(\Rightarrow1⋮d\Rightarrow d=1\)

\(\Rightarrow n+1\) và \(2n+3\) nguyên tố cùng nhau với mọi \(n\in N\)

Các câu sau em biến đổi tương tự

Do \(n\in N\Rightarrow2n+3\ge3\)

\(4n+23⋮2n+3\)

\(\Rightarrow4n+6+17⋮2n+3\)

Do \(4n+6=2\left(2n+3\right)⋮2n+3\)

\(\Rightarrow17⋮2n+3\)

\(\Rightarrow2n+3=Ư\left(17\right)=\left\{17\right\}\)

\(\Rightarrow2n+3=17\)

\(\Rightarrow n=7\)