Rút gọn biểu thức P .

1) Với \(a \ge 0,\,\,a \ne 9,\) ta có:

\(P = \frac{{2\sqrt a }}{{\sqrt a + 3}} + \frac{{\sqrt a + 1}}{{\sqrt a - 3}} + \frac{{ - 3 - 7\sqrt a }}{{\left( {\sqrt a + 3} \right)\left( {\sqrt a - 3} \right)}}\)

 \[ = \frac{{2\sqrt a \cdot \left( {\sqrt a - 3} \right)}}{{\left( {\sqrt a + 3} \right)\left( {\sqrt a - 3} \right)}} + \frac{{\left( {\sqrt a + 1} \right)\left( {\sqrt a + 3} \right)}}{{\left( {\sqrt a + 3} \right)\left( {\sqrt a - 3} \right)}} + \frac{{ - 3 - 7\sqrt a }}{{\left( {\sqrt a + 3} \right)\left( {\sqrt a - 3} \right)}}\]

 \[ = \frac{{2a - 6\sqrt a + a + 3\sqrt a + \sqrt a + 3 - 3 - 7\sqrt a }}{{\left( {\sqrt a + 3} \right)\left( {\sqrt a - 3} \right)}}\]

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

Vậy với \(a \ge 0,\,\,a \ne 9\) thì \(P = \frac{{3\sqrt a }}{{\sqrt a + 3}}.\)