Rút gọn biểu thức P .

a) Với \(x \ge 0,x \ne 1,\) ta có:

\(P = \frac{{\sqrt x }}{{\sqrt x - 1}} + \frac{3}{{\sqrt x + 1}} - \frac{{6\sqrt x - 4}}{{x - 1}}\)

 \( = \frac{{\sqrt x }}{{\sqrt x - 1}} + \frac{3}{{\sqrt x + 1}} - \frac{{6\sqrt x - 4}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\)

 \( = \frac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} + \frac{{3\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}} - \frac{{6\sqrt x - 4}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\)

 \( = \frac{{x + \sqrt x + 3\sqrt x - 3 - 6\sqrt x + 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\)

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

Vậy khi \(x \ge 0,\,\,x \ne 1\) thì \(P = \frac{{\sqrt x - 1}}{{\sqrt x + 1}}.\)