Giải thích các bước giải:
a) ĐKXĐ: $x>0$
Ta có:
$\begin{array}{l}
A = \dfrac{{{x^2} + \sqrt x }}{{x - \sqrt x + 1}} + 1 - \dfrac{{2x + \sqrt x }}{{\sqrt x }}\\
= \dfrac{{\sqrt x \left( {{{\left( {\sqrt x } \right)}^3} + 1} \right)}}{{x - \sqrt x + 1}} + 1 - 2\sqrt x - 1\\
= \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)\left( {x - \sqrt x + 1} \right)}}{{x - \sqrt x + 1}} - 2\sqrt x \\
= \sqrt x \left( {\sqrt x + 1} \right) - 2\sqrt x \\
= x - \sqrt x
\end{array}$
b) Ta có:
$\begin{array}{l}
A = x - \sqrt x \\
= \sqrt x \left( {\sqrt x - 1} \right)\\
< 0,\forall 0 < x < 1\\
\Rightarrow \left| A \right| = - A,\forall 0 < x < 1\\
\Rightarrow \left| A \right| + A = 0,\forall 0 < x < 1
\end{array}$
