\(b,n^2+5\) chia hết cho \(n+1\)
Ta có: \(n^2+5\) chia hết cho \(n+1\Leftrightarrow\)\(\frac{n^2+5}{n+1}\in Z\)
\(\frac{n^2+5}{n+1}=\frac{n^2+2x+1-2x+4}{n+1}=\frac{\left(n+1\right)^2}{n+1}+\frac{4-2x}{n+1}=n+1+\frac{-2x-2}{n+1}+\frac{6}{n+1}=n-1+\frac{6}{n+1}\)
Khi \(\frac{n^2+5}{n+1}\in Z\Rightarrow\left\{{}\begin{matrix}n\in Z\\\frac{6}{n+1}\in Z\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}n\in Z\\\left(n+1\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\end{matrix}\right.\left(1\right)\)
Từ \(\left(1\right)\) ta có thể suy ra được \(n\in\left\{0;\pm2;1;-3;-4;5;-7\right\}\)
Vậy ..................................
ta có : \(3n+25⋮n-4\)
\(\Rightarrow\frac{3n+25}{n-4}\in Z\) \(\Rightarrow3+\frac{37}{n-4}\in Z\)
vì \(3\in Z\) nên để \(\frac{3n+25}{n-4}\in Z\)
thì \(\frac{37}{n-4}\in Z\Leftrightarrow n-4\inƯ\left(37\right)=\left\{\pm1;\pm37\right\}\)
ta có bảng giá trị:
| n-4 | 1 | -1 | 37 | -37 |
| n | 5(tm | 3(TM) | 41(tm) | -33(tm) |
vậy ...
