Đáp án:
\(\dfrac{{2a + 2\sqrt a + 2}}{{\sqrt a }}\)
Giải thích các bước giải:
\(\begin{array}{l}
P = \dfrac{{\left( {\sqrt a - 1} \right)\left( {a + \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a - 1} \right)}} - \dfrac{{\left( {\sqrt a + 1} \right)\left( {a - \sqrt a + 1} \right)}}{{\sqrt a \left( {\sqrt a + 1} \right)}} + \left( {\dfrac{{a - 1}}{{\sqrt a }}} \right).\left[ {\dfrac{{a + 2\sqrt a + 1 + a - 2\sqrt a + 1}}{{a - 1}}} \right]\\
= \dfrac{{a + \sqrt a + 1}}{{\sqrt a }} - \dfrac{{a - \sqrt a + 1}}{{\sqrt a }} + \dfrac{{2a + 2}}{{\sqrt a }}\\
= \dfrac{{a + \sqrt a + 1 - a + \sqrt a - 1 + 2a + 2}}{{\sqrt a }}\\
= \dfrac{{2a + 2\sqrt a + 2}}{{\sqrt a }}
\end{array}\)
