$\displaystyle \begin{array}{{>{\displaystyle}l}} A=\frac{-6}{2\sqrt{x} -6} \ và\ B=\frac{2\sqrt{x}}{\sqrt{x} +3} +\frac{\sqrt{x}}{\sqrt{x} -3} +\frac{3x-3}{9-x} \ \\ a) \ x=3-2\sqrt{2} =2-2\sqrt{2} +1=\left(\sqrt{2} -1\right)^{2} \ \\ \rightarrow \sqrt{x} =|\sqrt{2} -1|=\sqrt{2} -1\ \\ A=\frac{-6}{2\sqrt{x} -6} =\frac{-3}{\sqrt{x} -3} \ \\ A=\frac{-3}{\sqrt{2} -1-3} =\frac{-3}{\sqrt{2} -4} \ \\ b) \ B=\frac{2\sqrt{x}\left(\sqrt{x} -3\right) +\sqrt{x}\left(\sqrt{x} +3\right) -3x+3}{x-9} \ \\ B=\frac{2x-6\sqrt{x} +x+3\sqrt{x} -3x+3}{x-9}\\ B=\frac{-3\sqrt{x} +3}{x-9} =\frac{-3\left(\sqrt{x} -1\right)}{x-9} \ \\ M=\frac{B}{A} =\frac{-3}{\sqrt{x} +3} .\frac{x-9}{-3\left(\sqrt{x} -1\right)} =\frac{\sqrt{x} -3}{\sqrt{x} -1} \ \\ .c) M=\frac{\sqrt{x} -3}{\sqrt{x} -1} =\frac{\sqrt{x} -1-2}{\sqrt{x} -1} =1-\frac{2}{\sqrt{x} -1} \ \\ M\ đạt\ giá\ trị\ nhỏ\ nhất\ khi\ 1-\frac{2}{\sqrt{x} -1} \ nhỏ\ nhất\ \\ \rightarrow \frac{-2}{\sqrt{x} -1} \ nhỏ\ nhất\ \rightarrow \ \sqrt{x} -1\ lớn\ nhất\ \\ ta\ có\ :\ \sqrt{x} -1\geqslant 1\ \\ Dấu\ =\ xảu\ ra\ khi\ x=0\ \\ Vậy\ minM=1-\frac{2}{0-1} =1-( -2) =3\ \\ \\ \\ \ \\ \ \\ \ \\ \ \end{array}$
