2) Chứng minh B= 2/ ( căn x + 2)

Với \(x \ge 0,\,x \ne 4,\) ta có:

\(B = \frac{1}{{\sqrt x - 2}} + \frac{{\sqrt x - 6}}{{x - 4}}\)\( = \frac{1}{{\sqrt x - 2}} + \frac{{\sqrt x - 6}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\)

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

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

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