`~rai~`
\(y=\sqrt{\dfrac{\sin3x+2}{1-\cos x}}\\ĐKXĐ:\\\begin{cases}\dfrac{\sin3x+2}{1-\cos x}\ge 0\\\cos x\ne 1\end{cases}\quad(1)\\\text{Ta có:}\\+)-1\le \sin3x\le 1\\\Leftrightarrow 1\le \sin3x+2\le 3\\\Leftrightarrow \sin3x+2>0\quad\forall x.(1)\\+)\cos x\le 1\\\Leftrightarrow 1-\cos x\ge 0\quad\forall x.(3)\\\text{Từ (2) và (3)}\Rightarrow \dfrac{\sin3x+2}{1-\cos x}>0\quad nên\\(1)\Leftrightarrow \cos x\ne 1\\\Leftrightarrow x\ne k2\pi.(k\in\mathbb{Z})\\TXĐ:D=\mathbb{R}\backslash\{k2\pi|k\in\mathbb{Z}\}.\\2.y=\dfrac{\sqrt{\cos4x-1}}{\sin2x-1}\\ĐKXĐ:\begin{cases}\cos4x-1\ge 0\\\sin2x-1\ne 0\end{cases}\\\Leftrightarrow \begin{cases}\cos4x\ge 1\\\sin2x\ne 1\end{cases}\\\Leftrightarrow \begin{cases}\cos4x=1\quad\text{(do }\cos4x\le 1)\\2x\ne\dfrac{\pi}{2}+k2\pi\end{cases}\\\Leftrightarrow \begin{cases}4x=k2\pi\\x=\dfrac{\pi}{4}+k\pi\end{cases}\\\Leftrightarrow \begin{cases}x=k\dfrac{\pi}{2}\\x\ne \dfrac{\pi}{4}+k\pi\end{cases}\\\Leftrightarrow x=k\dfrac{\pi}{2}.(k\in\mathbb{Z})\\D=\left\{k\dfrac{\pi}{2}\Big|k\in\mathbb{Z}\right\}.\)
