c) Rút gọn biểu thức B

c) Với \(x \ge 0;x \ne 1\), ta có:

\(B = \frac{{\sqrt x + 5}}{{\sqrt x + 1}} + \frac{{7 - \sqrt x }}{{x - 1}} = \frac{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} + \frac{{7 - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\)

   \( = \frac{{x + 4\sqrt x - 5 + 7 - \sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = \frac{{x + 3\sqrt x + 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\)

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

Vậy với \(x \ge 0;x \ne 1\) thì \(B = \frac{{\sqrt x + 2}}{{\sqrt x - 1}}\).