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) Tính GTNN của A.

a) \(A = \frac{{{a^2} + \sqrt a }}{{a - \sqrt a + 1}} - \frac{{2a + \sqrt a }}{{\sqrt a }} + 1\)

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

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

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

\( = a + \sqrt a - 2\sqrt a - 1 + 1\)\( = a - \sqrt a \).

b) Ta có: A = \(a - \sqrt a \)

\( = {\left( {\sqrt a } \right)^2} - 2\sqrt a .\frac{1}{2} + \frac{1}{4} - \frac{1}{4} = {\left( {\sqrt a - \frac{1}{2}} \right)^2} - \frac{1}{4}\)

Mà ta có: \({\left( {\sqrt a - \frac{1}{2}} \right)^2} \ge 0,\forall a \ge 0\)

Suy ra \(A \ge - \frac{1}{4}\)

Vậy \({A_{\min }} = \frac{{ - 1}}{4}\) khi \(a = \frac{1}{4}\).