\(\begin{array}{l}
1)\quad y = \sin\sqrt{\dfrac{1+x}{1-x}}\\
\text{Hàm số xác định}\\
\Leftrightarrow \dfrac{1+x}{1-x}\geqslant 0\\
\Leftrightarrow \left[\begin{array}{l}\begin{cases}1 + x \geqslant 0\\1 - x >0\end{cases}\\\begin{cases}1+ x \leqslant 0\\1 - x <0\end{cases}\end{array}\right.\\
\Leftrightarrow -1 \leqslant x < 1\\
\text{Vậy}\ D = [-1;1)\\
2)\quad y = \sqrt{\dfrac{2 + \cos x}{1 + \cos x}}\\
\text{Hàm số xác định}\\
\Leftrightarrow 1 + \cos  x\ne 0\\
\Leftrightarrow \cos x \ne - 1\\
\Leftrightarrow x \ne \pi + k2\pi\quad (k\in\Bbb Z)\\
\text{Vậy}\ D = \Bbb R\backslash\{\pi + k2\pi\ \Big|\ k\in\Bbb Z\}\\
3)\quad y = \dfrac{\cot x}{\cos x - 1}\\
\text{Hàm số xác định}\\
\Leftrightarrow \begin{cases}\sin x \ne 0\\\cos x \ne 1\end{cases}\\
\Leftrightarrow \begin{cases}x \ne k\pi\\x \ne k2\pi\end{cases}\\
\Leftrightarrow x \ne k\pi\quad (k\in\Bbb Z)\\
\text{Vậy}\ D = \Bbb R\backslash\{k\pi\ \Big|\ k\in\Bbb Z\}
\end{array}\)

`~rai~`

\(1/y=sin\sqrt{\dfrac{1+x}{1-x}}\\ĐKXĐ:\dfrac{1+x}{1-x}\ge 0\\\Leftrightarrow -1\le x<1.\\\text{(bước này bạn có thể kẻ bảng hoặc trục để xét dấu ra nháp nhé.)}\\TXĐ:D=[-1;1)\\2/y=\sqrt{\dfrac{2+sinx}{1+cosx}}\\ĐKXĐ:\begin{cases}\dfrac{2+sinx}{1+cosx}\ge 0\\cosx\ne -1\end{cases}.\quad (1)\\\text{Ta có:}\\+)-1\le sinx\le 1\\\Leftrightarrow 1\le 2+sinx\le 3\\\Rightarrow 2+sinx>0\quad\forall x\in\mathbb{R}.(2)\\+)-1\le cosx\le 1\\\Leftrightarrow 0\le 1+cosx\le 2\\\Rightarrow 1+cosx\ge 0\quad\forall x\in\mathbb{R}.(3)\\\text{Từ (2) và (3)}\Rightarrow \dfrac{2+sinx}{1+cosx}>0\quad\forall x\in\mathbb{R}\quad nên\\(1)\Leftrightarrow cosx\ne -1\\\Leftrightarrow x\ne \pi+k2\pi.(k\in\mathbb{Z}).\\TXĐ:D=mathbb{R}\backslash\{\pi+k2\pi|k\in\mathbb{Z}\}.\\3/y=\dfrac{\cot{x}}{\cos{x}-1}\\\Leftrightarrow \begin{cases}\sin{x}\ne 0\\cosx\ne 1\end{cases}\\\Leftrightarrow \begin{cases}x\ne k\pi\\x\ne k2\pi\end{cases}\\\Leftrightarrow x\ne k\pi.(k\in\mathbb{Z}).\\TXĐ:D=\mathbb{R}\backslash\{k\pi|k\in\mathbb{Z}\}\)