c) Rút gọn biểu thức B

c) Với \(x > 0,x \ne 1\), ta có:

\(B = \frac{1}{{x + \sqrt x }} + \frac{{2\sqrt x }}{{x - 1}} - \frac{1}{{x - \sqrt x }}\)

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

\[ = \frac{{\sqrt x - 1}}{{\sqrt x \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}} + \frac{{2\sqrt x \cdot \sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}} - \frac{{\sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\]

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

\( = \frac{{2\left( {x - 1} \right)}}{{\sqrt x \left( {x - 1} \right)}}\)\( = \frac{2}{{\sqrt x }}\).

Vậy với \(x > 0,x \ne 1\) thì \(B = \frac{2}{{\sqrt x }}\).