Đáp án:
$1)\\ a) x \ge -4\\ b)x >9\\ c)x >-3$
Giải thích các bước giải:
$1)\\ a)A=\sqrt{x+4}\\ ĐKXĐ: x+4 \ge 0 \Leftrightarrow x \ge -4\\ b)B=\dfrac{1}{\sqrt{x-9}}\\ ĐKXĐ: \left\{\begin{array}{l} x-9 \ge 0 \\ \sqrt{x-9} \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge 9 \\ x \ne 9 \end{array} \right.\\ \Leftrightarrow x >9\\ c)\sqrt{x+3}+\dfrac{1}{\sqrt{x+3}}\\ ĐKXĐ: \left\{\begin{array}{l} x+3 \ge 0 \\ \sqrt{x+3} \ne 0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} x \ge -3 \\ x \ne -3 \end{array} \right.\\ \Leftrightarrow x >-3\\ 2)\\ a)VT=2\sqrt{2+\sqrt{3}}\\ =\sqrt{8+4\sqrt{3}}\\ =\sqrt{2+2.\sqrt{2}.\sqrt{2}.\sqrt{3}+6}\\ =\sqrt{2+2.\sqrt{2}.\sqrt{6}+6}\\ =\sqrt{(\sqrt{2}+\sqrt{6})^2}\\ =\sqrt{2}+\sqrt{6}\\ =VP\\ b)VT=\sqrt{1+\dfrac{\sqrt{3}}{2}}\\ =\sqrt{\dfrac{2+\sqrt{3}}{2}}\\ =\sqrt{\dfrac{4+2\sqrt{3}}{4}}\\ =\sqrt{\dfrac{3+2\sqrt{3}+1}{4}}\\ =\sqrt{\dfrac{(\sqrt{3}+1)^2}{2^2}}\\ =\dfrac{\sqrt{3}+1}{2}\\ =VP$
