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a: \(\Leftrightarrow2n^2+n-2n-1+3⋮2n+1\)
\(\Leftrightarrow2n+1\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{0;-1;1;-2\right\}\)
b: \(\Leftrightarrow2n^2-4n+5n-10+3⋮n-2\)
\(\Leftrightarrow n-2\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{3;1;5;-1\right\}\)
c: \(\Leftrightarrow10n^2-15n+8n-12+7⋮2n-3\)
\(\Leftrightarrow2n-3\in\left\{1;-1;7;-7\right\}\)
hay \(n\in\left\{2;1;5;-2\right\}\)
d: \(\Leftrightarrow2n^2-n+4n-2+5⋮2n-1\)
\(\Leftrightarrow2n-1\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{1;0;3;-2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét 2n^2 +3n+3
= 2n^2 -n + 4n-2+5
= n.(2n-1)+2(2n-1)+5
= (2n-1).(n+2) + 5
để (2n-1).(n+2) + 5 chia hết cho 2n + 1 mà (2n-1).(n+2) chia hết cho 2n + 1
=> 5 chia hết cho 2n-1
=> 2n-1 thuộc ước 5={ +-1,+-5}
=> n thuộc { 0;1;-2;3}
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(3n^3-2n+17=3n^3+6n^2-6n^2-12n+10n+20-3\)
\(=3n^2\left(n+2\right)-6n\left(n+2\right)+10\left(n+2\right)-3\)
\(=\left(3n^2-6n+10\right)\left(n+2\right)-3\)
Để \(3n^3-2n+17⋮\left(n+2\right)\)
Hay \(\left(3n^2-6n+10\right)\left(n+2\right)-3⋮\left(n+2\right)\)
Mà \(\left(3n^2-6n+10\right)\left(n+2\right)⋮\left(n+2\right)\)
Suy ra \(3⋮\left(n+2\right)\)
Vì \(n\in Z\), suy ra \(n+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
TH1: \(n+2=\pm1\Leftrightarrow\left[{}\begin{matrix}n+2=1\\n+2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=-1\\n=-3\end{matrix}\right.\)
TH2:\(n+2=\pm3\Leftrightarrow\left[{}\begin{matrix}n+2=3\\n+2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=1\\n=-5\end{matrix}\right.\)
Vậy \(n\in\left\{-5;-3;-1;1\right\}\)
bài cô hoa chi
dan fa đáp án mấy , minhf làm hết rồi còn đáp án xem cho chắc chứ