Cho biểu thức: \(A = \frac{{{a^2} + \sqrt a }}{{a - \sqrt a + 1}} - \frac{{2a + \sqrt a }}{{\sqrt a }} + 1\).

a) Rút gọn A.

b) Tìm a để A = 2.

a) Ta có: \(A = \frac{{{a^2} + \sqrt a }}{{a - \sqrt a + 1}} - \frac{{2a + \sqrt a }}{{\sqrt a }} + 1\)
\( = \frac{{\sqrt a \left( {{{\left( {\sqrt a } \right)}^3} + 1} \right)}}{{a - \sqrt a + 1}} - \frac{{\sqrt a \left( {2\sqrt a + 1} \right)}}{{\sqrt a }} + 1\)

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

\( = \sqrt a \left( {\sqrt a + 1} \right) - 2\sqrt a \)

\( = a + \sqrt a - 2\sqrt a = a - \sqrt a \)

b) ĐKXĐ: a ≥ 0.

Để A = 2 thì: \(a - \sqrt a = 2\)

\( \Leftrightarrow a - \sqrt a - 2 = 0\)

\( \Leftrightarrow \left( {a + \sqrt a } \right) - \left( {2\sqrt a + 2} \right) = 0\)

\( \Leftrightarrow \sqrt a \left( {\sqrt a + 1} \right) - 2\left( {\sqrt a + 1} \right) = 0\)

\( \Leftrightarrow \left( {\sqrt a + 1} \right)\left( {\sqrt a - 2} \right) = 0\)

\( \Leftrightarrow \sqrt a - 2 = 0\)

\( \Leftrightarrow \)a = 4 (TMĐK)

Vậy với a = 4 thì A = 2.