Đáp án đúng: A
Giải chi tiết:Điều kiện: \(x > 0\,\,;\,\,x \ne 9\)
\(\begin{array}{l}M = A:B = \frac{6}{{\sqrt x \left( {\sqrt x - 3} \right)}}:\left( {\frac{{2\sqrt x }}{{x - 9}} - \frac{2}{{\sqrt x + 3}}} \right)\\\,\,\,\,\,\,\, = \frac{6}{{\sqrt x \left( {\sqrt x - 3} \right)}}:\frac{{2\sqrt x - 2\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\\,\,\,\,\,\,\, = \frac{6}{{\sqrt x \left( {\sqrt x - 3} \right)}}:\frac{6}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}\\\,\,\,\,\,\,\, = \frac{6}{{\sqrt x \left( {\sqrt x - 3} \right)}}.\frac{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}}{6}\\\,\,\,\,\,\,\, = \frac{{\sqrt x + 3}}{{\sqrt x }}\end{array}\)
Vậy \(M = \frac{{\sqrt x + 3}}{{\sqrt x }}.\)
Chọn A.