Câu 2:
$\sqrt{x+2}-\sqrt{x} > x+\dfrac{3}{4}$ (*)
Đk: $x \ge 0$
(*) $\rightarrow \sqrt{x+2}-\dfrac{3}{2}+\dfrac{1}{2}-\sqrt{x}+\dfrac{1}{4}-x > 0$
$\leftrightarrow \dfrac{x-\dfrac{1}{4}}{\sqrt{x+2}+\dfrac{3}{2}} + \dfrac{\dfrac{1}{4}-x}{\sqrt{x}+\dfrac{1}{2}}+\dfrac{1}{4}-x > 0$
$\leftrightarrow (x-\dfrac{1}{4})(\dfrac{1}{\sqrt{x+2} + \dfrac{3}{2}}-\dfrac{1}{\sqrt{x}+\dfrac{1}{2}}-1) > 0$
Ta có: $\sqrt{x+2}+\dfrac{3}{2} > \sqrt{x}+\dfrac{1}{2}$
$\rightarrow \dfrac{1}{\sqrt{x+2}+\dfrac{3}{2}} < \dfrac{1}{\sqrt{x}+\dfrac{1}{2}}$
$\rightarrow \dfrac{1}{\sqrt{x+2}+\dfrac{3}{2}}-\dfrac{1}{\sqrt{x}+\dfrac{1}{2}}-1 < 0$
$\rightarrow x < \dfrac{1}{4}$
Kết hợp với điều kiện ta có: $S = [0;\dfrac{1}{4})$