1) Đk: $\left\{ \begin{array}{l} x^2+x-2\ge0 \\ \sqrt{x^2+x-2}-x>0 \end{array} \right .\Leftrightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} x\ge1\\ x\le-2 \end{array} \right .(1) \\ \sqrt{x^2+x-2}>x \text{ (2)}\end{array} \right .$ Giải bất phương trình (2) TH1: $\left\{ \begin{array}{l} xx^2\end{array} \right .\Rightarrow \left\{ \begin{array}{l} x\ge0\\x>2 \end{array} \right .\Leftrightarrow x>2$ Kết hợp với (1) Vậy với $x>2$ và $x\le-2$ hàm số xác định TXĐ: $D=(-\infty;-2]\cup(2;+\infty)$ 2) Đk: $|1-\sin x|>0\Leftrightarrow 1-\sin x\ne0$ $\Leftrightarrow \sin x\ne1$ $\Leftrightarrow x\ne\dfrac{\pi}{2}+k2\pi(k\in\mathbb Z)$ Vậy TXĐ: $D=\mathbb R\backslash\{\dfrac{\pi}{2}+k2\pi,k\in\mathbb Z\}$ 4) Đk: $\left\{ \begin{array}{l} x+1\ge0 \\ 5-x>0\\\ln(5-x)\ne0\end{array} \right .\Leftrightarrow \left\{ \begin{array}{l} x\ge-1 \\ x0 \\ x+1>0\end{array} \right .\Leftrightarrow \left\{ \begin{array}{l} x1\end{array} \right .$ $\Leftrightarrow 10 \\\log_{0,8}\dfrac{2x+1}{x+5}-1\ge0 \end{array} \right .\Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l} x\dfrac{-1}{2}\end{array} \right .\\ \log_{0,8}\dfrac{2x+1}{x+5}\ge1\end{array} \right .$ $\Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l} x\dfrac{-1}{2}\end{array} \right .\\ \dfrac{2x+1}{x+5}\le 0,8\end{array} \right . $