Tìm TXĐ cho các hàm số

Tìm TXĐ cho các hàm số

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1. Đáp án:

$\begin{array}{l} a)Dkxd:\left\{ \begin{array}{l} x + 5 \ge 0\\ 5 – 2x \ge 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x \ge – 5\\ x \le \dfrac{5}{2} \end{array} \right.\\ Vay\, – 5 \le x \le \dfrac{5}{2}\\ b)Dkxd:\left\{ \begin{array}{l} {x^2} – 9 \ne 0\\ 2x + 3 \ge 0 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} x \ne 3;x \ne – 3\\ x \ge – \dfrac{3}{2} \end{array} \right.\\ Vay\,x \ge – \dfrac{3}{2};x \ne 3\\ c)Dkxd:2x + 6 > 0\\ \Rightarrow x > – 3\\ Vay\,x > – 3\\ d)\left\{ \begin{array}{l} 4 – x \ge 0\\ x > 0 \end{array} \right. \Rightarrow 0 < x \le 4\\ e)x + 2 \ne 0\\ \Rightarrow x \ne – 2\\ \Rightarrow TXD:D = R\backslash \left\{ { – 2} \right\}\\ f)Dkxd:\left\{ \begin{array}{l} x + 1 \ge 0\\ 1 – x \ge 0 \end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l} x \ge – 1\\ x \le 1 \end{array} \right.\\ \Rightarrow – 1 \le x \le 1\\ \Rightarrow TXD:D = \left[ { – 1;1} \right] \end{array}$