Bài 1:
\(2n+3\vdots n-2\)
\(2(n-2)+7\vdots n-2\)
\(7\vdots n-2\)
\(\Rightarrow n-2\in \text{Ư(7)}\Rightarrow n-2\in\left\{\pm 1;\pm 7\right\}\)
\(\Rightarrow n\in \left\{1;3;-5;9\right\}\)
Mà $n$ là số tự nhiên nên $n=1,3,9$
Bài 2:
\(3n+1\vdots 1-2n\)
\(\Rightarrow 2(3n+1)\vdots 1-2n\)
\(\Rightarrow 6n+2\vdots 1-2n\)
\(\Rightarrow 5-3(1-2n)\vdots 1-2n\)
\(\Rightarrow 5\vdots 1-2n\Rightarrow 1-2n\in\left\{\pm 1;\pm 5\right\}\)
\(\Rightarrow n\in\left\{0; 1;3; -2\right\}\)
Vì $n$ là số tự nhiên nên $n=0,1,3$
Bài 3:
Có: \(143=1.143=143.1=11.13=13.11=(-1).(-143)=....\)
Với $x\in\mathbb{N}$ thì $x+1\in\mathbb{N}^*$ nên ta xét các TH sau:
TH1: \(\left\{\begin{matrix} x+1=1\\ 2y-5=143\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=0\\ y=74\end{matrix}\right.\)
TH2: \(\left\{\begin{matrix} x+1=143\\ 2y-5=1\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=142\\ y=3\end{matrix}\right.\)
TH3: \(\left\{\begin{matrix} x+1=11\\ 2y-5=13\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=10\\ y=9\end{matrix}\right.\)
TH4: \(\left\{\begin{matrix} x+1=13\\ 2y-5=11\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=12\\ y=8\end{matrix}\right.\)
Vậy.........
