Cho biểu thức \(A = 1 + \left( {\frac{{2a + \sqrt a - 1}}{{1 - a}} - \frac{{2a\sqrt a - \sqrt a + a}}{{1 - a\sqrt a }}} \right).\frac{{a - \sqrt a }}{{2\sqrt a - 1}}\). Rút gọn A.

Điều kiện: a ≥ 0; a ≠ 1; a ≠ \(\frac{1}{4}\)

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

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

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

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

\[A = 1 + \left( {2\sqrt a - 1} \right)\left( {\sqrt a + 1} \right).\frac{{1 - \sqrt a }}{{\left( {1 - a} \right)\left( {1 - a\sqrt a } \right)}}.\frac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{2\sqrt a - 1}}\]

\[A = 1 - \frac{{\sqrt a \left( {\sqrt a - 1} \right)}}{{a\sqrt a - 1}}\]

\[A = \frac{{a\sqrt a - 1 - a + \sqrt a }}{{a\sqrt a - 1}}\]

\[A = \frac{{\sqrt a \left( {a + 1} \right) - \left( {a + 1} \right)}}{{a\sqrt a - 1}} = \frac{{\left( {a + 1} \right)\left( {\sqrt a - 1} \right)}}{{a\sqrt a - 1}}\].