Question

tìm số nguyên n để

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

Answer 1

\(n^2+3n+3⋮2n-1\)

Mà \(2n-1⋮2n-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}2n^2+6n+6⋮2n-1\\2n^2-n⋮2n-1\end{matrix}\right.\)

\(\Leftrightarrow7n+6⋮2n-1\)

Mà \(2n-1⋮2n-1\)

\(\Leftrightarrow\left\{{}\begin{matrix}14n+12⋮2n-1\\14n-7⋮2n-1\end{matrix}\right.\)

\(\Leftrightarrow19⋮2n-1\)

\(\Leftrightarrow2n-1\inƯ\left(19\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2n-1=1\\2n-1=19\\2n-1=-1\\2n-1=-19\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=1\\n=10\\n=0\\n=-9\end{matrix}\right.\)

Vậy ..

\(\)

Answer 2

Ta có:

(1² +3*1+3)=1+3+3=7

(2*1-1)=2-1=1

7:1=7

Suy ra n=1

Answer 3

Ta có:

\(n^2+3n+3=2n^2+6n+6=n\left(2n-1\right)+3\left(2n-1\right)+n+9\)

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

Nên để \(n^2+3n+3⋮2n-1\)

Thì n+9 \(⋮2n-1\)

=> 2n+18\(⋮2n-1\)

<=> 2n-1+19\(⋮2n-1\)

mà 2n-1\(⋮2n-1\)

Nên 19\(⋮2n-1\)

=> \(\left(2n-1\right)\inƯ\left(19\right)\)

=> n\(\in\left\{1;0;10;-9\right\}\)

Answer 4

xong rồi bạn chia trường hợp cho vô thử rồi loại từ từ ra