Đáp án:

$2)\\ a)x>2\\ b)x \ge 2\\ c)x>2\\ d)x<\dfrac{3}{2}\\ e)x>\dfrac{-3}{2}\\ f)x<-1\\ 3)\\ a)b)c) \forall \ x\\ d)x=1\\ e)x=5\\ f) x \in \varnothing\\ 4)\\ a)-2\le x \le 2\\ b) \left[\begin{array}{l} x \ge 4 \\ x \le -4\end{array} \right.\\ c) \left[\begin{array}{l} x \ge \sqrt{3} \\ x \le -\sqrt{3} \end{array} \right.\\ d) \left[\begin{array}{l} x \ge 3 \\ x \le -1 \end{array} \right.\\ e) \left[\begin{array}{l} x \ge 2 \\ x \le 0 \end{array} \right.\\ f) \left[\begin{array}{l} x \ge 3 \\ x \le 2 \end{array} \right.\\ 5)\\ a) \left[\begin{array}{l} x \ge 1 \\ x \le -1 \end{array} \right. \\ b)\left[\begin{array}{l} x \ge 4 \\ x \le -2 \end{array} \right. \\ c) -4 \le x \le 4\\ d) x \ge 1\\ e) x \ne \dfrac{3}{2}\\ f) x \ge 1$

Giải thích các bước giải:

$2)\\ a) ĐKXĐ: \left\{\begin{array}{l} x-2 \ge 0\\ x -2 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 2\\ x  \ne 2\end{array} \right.\\ \Leftrightarrow x>2\\ b)ĐKXĐ: \left\{\begin{array}{l} x-2 \ge 0\\ x +2 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 2\\ x  \ne -2\end{array} \right.\\ \Leftrightarrow x \ge 2\\ c)ĐKXĐ: \left\{\begin{array}{l} x-2 \ge 0\\x^2-4 \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 2\\ (x-2)(x+2) \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 2\\ x \ne 2 \\x \ne -2\end{array} \right.\\ \Leftrightarrow x>2\\ d)ĐKXĐ:  \left\{\begin{array}{l} \dfrac{1}{3-2x} \ge 0 \\ 3-2x \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 3-2x \ge 0 \\ 3-2x \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x\le \dfrac{3}{2} \\ x \ne \dfrac{3}{2} \end{array} \right.\\ \Leftrightarrow x<\dfrac{3}{2}\\ e)ĐKXĐ:  \left\{\begin{array}{l} \dfrac{4}{2x+3} \ge 0 \\ 2x+3 \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} 2x+3 \ge 0 \\ 2x+3 \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge \dfrac{-3}{2} \\ x \ne \dfrac{-3}{2} \end{array} \right.\\ \Leftrightarrow x>\dfrac{-3}{2}\\ f)ĐKXĐ:  \left\{\begin{array}{l} \dfrac{-2}{x+1} \ge 0 \\ x+1 \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x+1 \le 0 \\ x+1 \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \le - 1\\ x \ne -1 \end{array} \right.\\ \Leftrightarrow x<-1\\ 3)\\ a) x^2+1 >0 \ \forall \ x$

$\Rightarrow \sqrt{x^2+1}$ xác định với $\forall \ x$

$b)4x^2+3 >0 \ \forall \ x$

$\Rightarrow \sqrt{4x^2+3}$ xác định với $\forall \ x$

$c)9x^2-6x+1\\ =(3x)^2-2.3x.1+1\\ =(3x-1)^2 \ge 0 \ \forall \ x$

$\Rightarrow \sqrt{9x^2-6x+1}$ xác định với $\forall \ x$

$d)-x^2+2x-1\\ =-(x^2-2x+1)\\ =-(x-1)^2 \le 0 \ \forall \ x$

$\Rightarrow \sqrt{-x^2+2x-1}$ chỉ xác định khi $-(x-1)^2=0 \Leftrightarrow x=1$

$e)-|x-5| \le 0 \ \forall \ x$

$\Rightarrow \sqrt{-|x-5| }$  chỉ xác định khi $-|x-5| =0 \Leftrightarrow x=5$

$f)-2x^2 -1 < 0 \ \forall \ x$

$\Rightarrow \sqrt{-2x^2 -1 } $ luôn không xác định với $\forall x$

$4)\\ a)ĐKXĐ: 4-x^2 \ge 0\\ \Leftrightarrow (2-x)(2+x) \ge 0\\ \Leftrightarrow -2\le x \le 2\\ b)ĐKXĐ: x^2-16 \ge 0\\ \Leftrightarrow (x-4)(x+4) \ge 0\\ \Leftrightarrow  \left[\begin{array}{l} x \ge 4 \\ x \le -4\end{array} \right.\\ c)ĐKXĐ: x^2-3 \ge 0\\ \Leftrightarrow (x-\sqrt{3})(x+\sqrt{3}) \ge 0\\ \Leftrightarrow  \left[\begin{array}{l} x \ge \sqrt{3} \\ x \le -\sqrt{3} \end{array} \right.\\ d)ĐKXĐ: x^2-2x-3 \ge 0\\ \Leftrightarrow x^2-2x+1-4 \ge 0\\ \Leftrightarrow (x-1)^2-4 \ge 0\\ \Leftrightarrow (x-1-2)(x-1+2) \ge 0\\ \Leftrightarrow (x-3)(x+1) \ge 0\\ \Leftrightarrow  \left[\begin{array}{l} x \ge 3 \\ x \le -1 \end{array} \right.\\ e)ĐKXĐ: x(x-2) \ge 0\\ \Leftrightarrow  \left[\begin{array}{l} x \ge 2 \\ x \le 0 \end{array} \right.\\ f)ĐKXĐ: x^2-5x+6 \ge 0\\ \Leftrightarrow x^2-2x-3x+6 \ge 0\\ \Leftrightarrow x(x-2)-3(x-2) \ge 0\\ \Leftrightarrow (x-3)(x-2) \ge 0\\ \Leftrightarrow  \left[\begin{array}{l} x \ge 3 \\ x \le 2 \end{array} \right.\\ 5)\\a)ĐKXĐ: |x|-1 \ge 0\\ \Leftrightarrow |x| \ge 1\\ \Leftrightarrow  \left[\begin{array}{l} x \ge 1 \\ x \le -1 \end{array} \right. \\ b)ĐKXĐ: |x-1|-3 \ge 0\\ \Leftrightarrow |x-1| \ge 3\\ \Leftrightarrow  \left[\begin{array}{l} x-1 \ge 3 \\ x-1 \le -3 \end{array} \right. \\ \Leftrightarrow  \left[\begin{array}{l} x \ge 4 \\ x \le -2 \end{array} \right. \\ c)ĐKXĐ: 4-|x| \ge 0\\ \Leftrightarrow |x| \le 4\\ \Leftrightarrow  -4 \le x \le 4\\ d)ĐKXĐ: \left\{\begin{array}{l} x-1 \ge 0 \\ x-2\sqrt{x-1} \ge 0 \end{array} \right. \\ \Leftrightarrow \left\{\begin{array}{l} x \ge 1 \\ x \ge 2\sqrt{x-1} \end{array} \right. \\ \Leftrightarrow \left\{\begin{array}{l} x \ge 1 \\ x^2 \ge 4(x-1) \end{array} \right. \\ \Leftrightarrow \left\{\begin{array}{l} x \ge 1 \\ x^2-4x+4 \ge 0 \end{array} \right. \\ \Leftrightarrow \left\{\begin{array}{l} x \ge 1 \\ (x-2)^2 \ge 0 \end{array} \right. \\ \Leftrightarrow x \ge 1\\ e)ĐKXĐ: 9-12x+4x^2 >0\\ \Leftrightarrow (2x)^2-2.2x.3+3^2 >0\\ \Leftrightarrow (2x-3)^2 >0 (*)$

Do $(2x-3)^2 \ge 0 \forall x$ nên để $(*)$ đúng thì $2x-3 \ne 0 \Leftrightarrow x \ne \dfrac{3}{2}$

$f)ĐKXĐ: \left\{\begin{array}{l} x-1 \ge 0 \\ x+2\sqrt{x-1} \ge 0 \end{array} \right. \\ \Leftrightarrow \left\{\begin{array}{l} x \ge 1 \\ x+2\sqrt{x-1} \ge 0(**) \end{array} \right. $

$\Leftrightarrow x \ge 1 ( Do\ \ x\ge 1; 2\sqrt{x-1} \ge 0 \ \forall x \ge 1$ nên $(**) $ luôn đúng)