$\displaystyle \begin{array}{{>{\displaystyle}l}} A=\sqrt{8} -2\sqrt{18} +5\sqrt{32} -\sqrt{\left(\sqrt{2} -1\right)^{2}}\\ A=2\sqrt{2} -3\sqrt{2} +5.4\sqrt{2} -\sqrt{2} +1\\ A=18\sqrt{2} +1\ \\ b)\frac{2}{\sqrt{3} -1} -\frac{1}{\sqrt{3} -2} +\frac{12}{\sqrt{3} +3} \ \\ \frac{2\left(\sqrt{3} +1\right)}{2} -\frac{1\left(\sqrt{3} +2\right)}{1} +\frac{12\left(\sqrt{3} -3\right)}{6}\\ =\sqrt{3} +1-\sqrt{3} -2+2\sqrt{3} -6\\ =2\sqrt{3} -5\ \\ 2.\ \\ \sqrt{x^{2} -6x+9} =7\ \\ \rightarrow \sqrt{x^{2} -2.3x+9} =7\ \\ \rightarrow \sqrt{( x-3)^{2}} =7\ \\ \rightarrow |x-3|=7\ \\ \rightarrow \left[ \begin{array}{l l} x-3=7( x\geqslant 3) & \\ 3-x=7\ ( x< 7) & \end{array} \right.\\ \rightarrow \left[ \begin{array}{l l} x=10 & \\ x=-4\ & \end{array} \right.\\ vậy....\ \\ b)\sqrt{4x-20} +3\sqrt{\frac{x-5}{9}} -\frac{1}{3}\sqrt{9x-45} =6\ \\ 2\sqrt{x-5} +\sqrt{x-5} -\sqrt{x} -5=6\ \\ 2\sqrt{x-5} =6\ \\ \rightarrow \sqrt{x-5} =3\ \\ DK:\ x\geqslant 5\ \\ x-5=9\ \\ \rightarrow x=14\ \\ \ \end{array}$
