Chứng minh rằng B = 12 căn x/( x − 9) .

c) Với \(x \ge 0;x \ne 9\), ta có:

\(B = \frac{{\sqrt x + 3}}{{\sqrt x - 3}} - \frac{{\sqrt x - 3}}{{\sqrt x + 3}}\)

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

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

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

    \( = \frac{{12\sqrt x }}{{\left( {\sqrt x - 3} \right)\left( {\sqrt x + 3} \right)}} = \frac{{12\sqrt x }}{{x - 9}}.\)

Vậy với \(x \ge 0;x \ne 9\) thì \(B = \frac{{12\sqrt x }}{{x - 9}}.\)