$\begin{array}{l}a)\,\,y= \dfrac{1}{x^2 +1} + \dfrac{x}{x^2 - 4}\\ \text{y xác định}\,\Leftrightarrow \begin{cases}x^2 + 1 \ne 0 \\x^2 - 4\ne 0\end{cases}\\ \Leftrightarrow x \ne \pm 2\\ \Rightarrow TXĐ: D = R \backslash\left\{\pm 2\right\}\\ Hoặc\,\,TXĐ: D = (-\infty;-2)\cup(-2;2)\cup(2;+\infty)\\ b)\,\,y = \dfrac{1}{x\sqrt{x + 1}}\\ \text{y xác định}\,\Leftrightarrow \begin{cases}x\sqrt{x + 1} \ne 0\\x+1 > 0\end{cases}\\ \Leftrightarrow \begin{cases}x \ne 0\\x > - 1\end{cases}\\ \Rightarrow TXĐ: D = (-1;0)\cup (0;+\infty)\\ c)\,\,y = \sqrt{4x + 2} + \dfrac{x}{\sqrt{-x +1}}\\ \text{y xác định}\,\Leftrightarrow \begin{cases}4x + 2 \geq 0\\-x + 1 > 0\end{cases}\\ \Leftrightarrow - \dfrac{1}{2} \leq x < 1\\ \Rightarrow TXĐ: D = \left[-\dfrac{1}{2};1\right)\\ d)\,\,y = \dfrac{1}{(x-1)\sqrt{x-3}}\\ \text{y xác định}\,\Leftrightarrow \begin{cases}x - 1 \ne 0\\x - 3 > 0\end{cases}\Leftrightarrow x > 4\\ \Rightarrow TXĐ: D = (3;+\infty)\end{array}$
