2) Chứng minh \(B = \frac{{\sqrt x }}{{\sqrt x + 2}}.\)

Với \(x > 0,\,\,x \ne 4,\) ta có:

\(B = \frac{{4 - 6\sqrt x }}{{x - 4}} + \frac{2}{{\sqrt x + 2}} - \frac{{\sqrt x }}{{2 - \sqrt x }}\)

 \( = \frac{{4 - 6\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} + \frac{{2\left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} + \frac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\)

 \( = \frac{{4 - 6\sqrt x + 2\left( {\sqrt x - 2} \right) + \sqrt x \left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\)

 \( = \frac{{4 - 6\sqrt x + 2\sqrt x - 4 + x + 2\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} = \frac{{x - 2\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\)

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

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