$\displaystyle \begin{array}{{>{\displaystyle}l}} a)\sqrt{9-4\sqrt{5}} -\sqrt{9+4\sqrt{5}}\\ =\sqrt{4-2.2\sqrt{5} +5} -\sqrt{4+2.2\sqrt{5} +5} \ \\ =\sqrt{\left( 2-\sqrt{5}\right)^{2}} -\sqrt{\left( 2+\sqrt{5}\right)^{2}}\\ =|2-\sqrt{5} |-2-\sqrt{5} \ \\ =\sqrt{5} -2-2-\sqrt{5} =-4\ \\ b)\sqrt{3+2\sqrt{2}} +\sqrt{\left(\sqrt{2} -2\right)^{2}}\\ =\sqrt{2+2\sqrt{2} +1} +|\sqrt{2} -2|\\ =\sqrt{\left(\sqrt{2} +1\right)^{2}} +2-\sqrt{2} \ \\ =\sqrt{2} +1+2-\sqrt{2} =3\ \\ c)\sqrt{4-\sqrt{15}} -\sqrt{4+\sqrt{15}} -\sqrt{6} \ \\ Ta\ có\ :\ A=\sqrt{4-\sqrt{15}} -\sqrt{4+\sqrt{15}} \ \\ A^{2} =4-\sqrt{15} +4+\sqrt{15} -2(\sqrt{\left( 4-\sqrt{15}\right)\left( 4+\sqrt{15}\right)} \ \\ A^{2} =8-2=6\ \\ \rightarrow A=\sqrt{6} \ \\ do\ đó\ :\ \sqrt{4-\sqrt{15}} -\sqrt{4+\sqrt{15}} -\sqrt{6} =0\ ( \ mới\ đúng\ ) \ \\ d)\sqrt{6+2\sqrt{4-2\sqrt{3}}} \ \\ \sqrt{6+2\sqrt{3-2\sqrt{3} +1}}\\ \sqrt{6+2\sqrt{\left(\sqrt{3} -1\right)^{2}}}\\ \sqrt{6+2\sqrt{3} -2} =\sqrt{4+2\sqrt{3}} =\sqrt{3+2\sqrt{2} +1} =\sqrt{\left(\sqrt{3} +1\right)^{2}}\\ =\sqrt{3} +1\ \\ Câu\ 2:\ \\ a)\sqrt{x-3} =2\ \\ \rightarrow DK:x\geqslant 3\ \\ x-3=4\ \\ \rightarrow x=7\ \\ b) \ \sqrt{x^{2} -6x+9} =5\ \\ \rightarrow \sqrt{( x-3)^{2}} =5\ \\ \rightarrow |x-3|=5\ \\ \rightarrow \left[ \begin{array}{l l} x-3=5 & ( x\geqslant 3)\\ 3-x=5 & ( x< 3) \end{array} \right.\rightarrow \left[ \begin{array}{l l} x=8( tm) & \\ x=-2\ ( tm) & \end{array} \right.\\ \rightarrow vậy\\ d)\sqrt{4( x+2)^{2}} =8\ \\ 2|x+2|=8\\ \rightarrow |x+2|=4\ \\ \rightarrow \left[ \begin{array}{l l} x+2=4.( x\geqslant -2) & \\ -x-2=4\ ( x< \ -2) & \end{array} \right.\rightarrow \left[ \begin{array}{l l} x=2\ ( tm) & \\ x=-6\ ( tm) & \end{array} \right.\\ Vậy....\ \\ e)\sqrt{x^{2} +x-1} =2-x\ \\ \rightarrow \left[ \begin{array}{l l} 2-x\geqslant 0 & \\ x^{2} +x-1=4-4x+x^{2} & \end{array} \right.\rightarrow \left[ \begin{array}{l l} x\leqslant 2 & \\ 5x=5 & \end{array} \right.\rightarrow x=1\ ( tm) \ \\ Vậy\ nghiệm\ pt\ là\ x=1\ \\ \end{array}$