Giải thích các bước giải:
1.Ta có hàm số xác định
$\to\begin{cases} x\ge 0\\ x+1\ge 0\\ \sqrt{x}-\sqrt{x+1}\ne 0\\ \sqrt{x}+\sqrt{x+1}\ne 0\\ (\sqrt{x+1}-1)^2-(\sqrt{x+1}+1)^2\ge 0\end{cases}$
$\to\begin{cases} x\ge 0\\ x\ge -1\\ \sqrt{x}\ne \sqrt{x+1}\\ \sqrt{x}\ne -\sqrt{x+1}\\ -4\sqrt{x+1}\ne 0\end{cases}$
$\to\begin{cases} x\ge 0\\ x\ge -1\\ \forall x\ge 0\\ \forall x\ge 0\\ x+1>0\end{cases}$
$\to x\ge 0$
$\to D_f=[0,+\infty)$
2.Ta có:
$f(x)=\dfrac{\dfrac{1}{\sqrt{x}+\sqrt{x+1}}-\dfrac{1}{\sqrt{x}-\sqrt{x+1}}}{(\sqrt{x+1}-1)^2-(\sqrt{x+1}+1)^2}$
$\to f(x)=\dfrac{\dfrac{\sqrt{x}-\sqrt{x+1}-(\sqrt{x}+\sqrt{x+1})}{(\sqrt{x}+\sqrt{x+1})(\sqrt{x}-\sqrt{x+1})}}{-4\sqrt{x+1}}$
$\to f(x)=\dfrac{\dfrac{-2\sqrt{x+1}}{x-(x+1)}}{-4\sqrt{x+1}}$
$\to f(x)=\dfrac{\dfrac{-2\sqrt{x+1}}{-1}}{-4\sqrt{x+1}}$
$\to f(x)=-\dfrac12$
