[tex]\bf E(x)=\dfrac{2x-4}{x+3}=\dfrac{2x+6-10}{x+3}=\dfrac{2(x+3)}{x+3}-\dfrac{10}{x+3}=2-\dfrac{10}{x+3} \in\mathbb{N} \Rightarrow\\ \\ \\ \Rightarrow \dfrac{10}{x+3}=2\ \ \ sau\ \ \ \dfrac{10}{x+3}=1\\ \\ \\ I)\ \ \dfrac{10}{x+3}=2 \Rightarrow x+3=5 \Rightarrow x=2\\ \\ \\ II)\ \ \dfrac{10}{x+3}=1 \Rightarrow x+3=10 \Rightarrow x=7[/tex]
