what? dễ mà :)
Để \(\dfrac{5}{\sqrt{2x+1}+2}\) \(\in Z\)
\(\Rightarrow\sqrt{2x+1}+2\inƯ\left(5\right)\Leftrightarrow\sqrt{2x+1}+2\in\left\{-5;-1;1;5\right\}\)
mà \(\sqrt{2x+1}+2\ge2\) \(\Rightarrow\sqrt{2x+1}+2=5\)
\(\Leftrightarrow\sqrt{2x+1}=3\)
\(\Leftrightarrow2x+1=9\Leftrightarrow2x=8\Leftrightarrow x=4\) (t/m)
Vậy x = 4 thì biểu thức \(\in Z\)