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: $\left\{ \begin{array}{l} x<0 \\ x^2+x-2\ge0 \end{array} \right .\Leftrightarrow \left\{ \begin{array}{l} x<0 \\ \left[ \begin{array}{l} x\ge1\\ x\le-2 \end{array} \right . \end{array} \right .\Leftrightarrow x\le-2$
TH2: $\left\{ \begin{array}{l} x\ge0\\x^2+x-2>x^2\end{array} \right .\Rightarrow \left\{ \begin{array}{l} x\ge0\\x>2 \end{array} \right .\Leftrightarrow x>2$
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: $\left\{ \begin{array}{l} x+1\ge0 \\ 5-x>0\\\ln(5-x)\ne0\end{array} \right .\Leftrightarrow \left\{ \begin{array}{l} x\ge-1 \\ x<5\\5-x\ne1\end{array} \right .\Leftrightarrow \left\{ \begin{array}{l} x\ge-1 \\ x<5\\x\ne4\end{array} \right .$
$\Leftrightarrow \left\{ \begin{array}{l} -1\le x\le5 \\ x\ne4\end{array} \right .$
5) Đk: $\left\{ \begin{array}{l} 2-x>0 \\ x+1>0\end{array} \right .\Leftrightarrow \left\{ \begin{array}{l} x<2 \\ x>1\end{array} \right .$
$\Leftrightarrow 10 \\\log_{0,8}\dfrac{2x+1}{x+5}-1\ge0 \end{array} \right .\Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l} x<-5\\x>\dfrac{-1}{2}\end{array} \right .\\ \log_{0,8}\dfrac{2x+1}{x+5}\ge1\end{array} \right .$
$\Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l} x<-5\\x>\dfrac{-1}{2}\end{array} \right .\\ \dfrac{2x+1}{x+5}\le 0,8\end{array} \right . $
