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 $a = \frac{1}{4}$.