Phần a ,
x + 3 chia hết cho x + 1
x - 1 chia hết cho x - 1
\(\Rightarrow x+3-\left(x-1\right)=4\text{ }⋮\text{ }x-1\)
\(x-1\in\left\{1\text{ };\text{ }-1\text{ };\text{ }2\text{ };\text{ }-2\text{ };\text{ }4\text{ };\text{ }-4\right\}\)
\(\Rightarrow x\in\left\{2\text{ };\text{ }0\text{ };\text{ }3\text{ };\text{ }-1\text{ };\text{ }5\text{ };\text{ }-3\right\}\)
Phần b,
\(\frac{4x+3}{2x+1}=\frac{2\left(2x+1\right)+1}{2x+1}=\frac{2\left(2x+1\right)}{2x+1}+\frac{1}{2x+1}=2+\frac{1}{2x+1}\in Z\)
\(\Rightarrow1\text{ }⋮\text{ }2x+1\)
\(\Rightarrow2x+1\in\left\{1\text{ };\text{ }-1\right\}\)
\(\Rightarrow x=0\)vì \(x\in N\)
Cảm ơn bạn Nguyễn Thị Thu Thủy rất nhiều !