Rút gọn biểu thức B

2) Với \(x \ge 0,\,\,x \ne 1,\) ta có:

\(B = \frac{{5\left( {x - \sqrt x } \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}} + \frac{{x + \sqrt x + 9}}{{\sqrt x + 3}}\)

 \( = \frac{{5\sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 3} \right)}} + \frac{{x + \sqrt x + 9}}{{\sqrt x + 3}}\)

 \( = \frac{{5\sqrt x }}{{\sqrt x + 3}} + \frac{{x + \sqrt x + 9}}{{\sqrt x + 3}}\)

 \( = \frac{{5\sqrt x + x + \sqrt x + 9}}{{\sqrt x + 3}}\)

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

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

 \( = \sqrt x + 3.\)

Vậy \(B = \sqrt x + 3\) với \(x \ge 0;x \ne 1.\)