$\begin{array}{l}1)\quad y = \log_3\dfrac{10 -x}{x^2 - 3x + 2}\\ \text{Hàm số xác định }\Leftrightarrow \dfrac{10 -x}{x^2- 3x + 2} >0\\ \Leftrightarrow \left[\begin{array}{l}\begin{cases} 10 - x>0\\x^2 - 3x + 2 >0 \end{cases}\\\begin{cases}10 - x < 0\\x^2 - 3x + 2 <0 \end{cases}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}\begin{cases} x < 10\\\left[\begin{array}{l}x > 2\\x < 1\end{array}\right. \end{cases}\\\begin{cases}x > 10\\1 < x <2 \end{cases}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}2 < x < 10\\x < 1 \end{array}\right.\\ \Rightarrow TXĐ: D = (-\infty;1)\cup (2;10)\\ 2)\quad y = \sqrt{\log_3(x-2) - 3}\\ \text{Hàm số xác định }\Leftrightarrow \begin{cases}x - 2 >0\\\log_3(x-2) - 3 \geq 0\end{cases}\\ \Leftrightarrow \begin{cases}x -2 >0\\\log_3(x-2) \geq 3\end{cases}\\ \Leftrightarrow x - 2 \geq 27\\ \Leftrightarrow x \geq 29\\ \Rightarrow TXĐ: D = [29;+\infty)\\ 3)\quad y = (x^2 + 2x).e^{-x}\\ \to y= \dfrac{x^2 + 2x}{e^x}\\ \to y' = \dfrac{(x^2 + 2x)'.e^x - (e^x)'.(x^2 + 2x)}{e^{2x}}\\ \to y ' = \dfrac{(2x + 2)e^x - e^x(x^2 + 2x)}{e^{2x}}\\ \to y' = \dfrac{2x + 2 - x^2 - 2x}{e^x}\\ \to y' =\dfrac{-x^2 + 2}{e^x}\\ \qquad\left[= (2 - x^2).e^{-x}\right] \end{array}$
