$\displaystyle \begin{array}{{>{\displaystyle}l}} Bài\ 3:\\ a.\ ĐK:\ x\geqslant 0\\ 2x=2\\ \Leftrightarrow x=1\ ( TM)\\ Vậy\ S=\{1\}\\ b.\ x^{2} -4x+4=9\\ \Leftrightarrow x^{2} -4x-5=0\\ \Leftrightarrow ( x-5)( x+1) =0\\ \Leftrightarrow x=5;\ x=-1\\ Vậy\ S=\{-1;5\}\\ c.\ ĐK\ x\geqslant -1\\ x+1=5+9+6\sqrt{5}\\ \Leftrightarrow x=13+6\sqrt{5} \ ( TM)\\ Vậy\ \ S=\left\{13+6\sqrt{5}\right\}\\ d.\ ĐK:\ x\geqslant 3\\ \Leftrightarrow 2x+1+4+4\sqrt{2x+1} =x-3\\ \Leftrightarrow 4\sqrt{2x+1} =-x-8\ ( x\leqslant -8)\\ \Leftrightarrow 16( 2x+1) =x^{2} +16x+64\\ \Leftrightarrow x^{2} -16x+48=0\\ \Leftrightarrow x=12\ ( loại;\ x=4\ ( TM)\\ Vậy\ S=\{4\}\\ Bài\ 4:\\ a.\ ĐKXĐ:\ x\geqslant 0;\ x\neq 1\\ A=\frac{\sqrt{x} -1-\sqrt{x} -1}{\left(\sqrt{x} -1\right)\left(\sqrt{x} +1\right)} .\frac{\sqrt{x}\left(\sqrt{x} +1\right)}{2}\\ =\frac{-2}{\left(\sqrt{x} -1\right)\left(\sqrt{x} +1\right)} .\frac{\sqrt{x}\left(\sqrt{x} +1\right)}{2}\\ =\frac{-\sqrt{x}}{\sqrt{x} -1}\\ b.\ A >0\\ \Leftrightarrow \frac{-\sqrt{x}}{\sqrt{x} -1} >0\Leftrightarrow \sqrt{x} -1< 0\Leftrightarrow 0\leqslant x< 1\\ Vậy\ S=\{x|0\leqslant x< 1\}\\ \end{array}$
