Rút gọn biểu thức B .

Ta có: \(B = \frac{3}{{\sqrt x  + 2}} - \frac{{\sqrt x }}{{2 - \sqrt x }} + \frac{{9\sqrt x  - 10}}{{4 - x}}\)

                  \( = \frac{{3\left( {\sqrt x  - 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}} + \frac{{\sqrt x \left( {\sqrt x  + 2} \right)}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}} - \frac{{9\sqrt x  - 10}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\)

                 \( = \frac{{3\sqrt x  - 6 + x + 2\sqrt x  - 9\sqrt x  + 10}}{{\left( {\sqrt x  + 2} \right)\left( {\sqrt x  - 2} \right)}}\)

                 \( = \frac{{x - 4\sqrt x  + 4}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\)

                 \( = \frac{{{{\left( {\sqrt x  - 2} \right)}^2}}}{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 2} \right)}}\)

                \( = \frac{{\sqrt x  - 2}}{{\sqrt x  + 2}}\).

Vậy \(B = \frac{{\sqrt x  - 2}}{{\sqrt x  + 2}}\) với \(x \ge 0\), \(x \ne 4\).