Giải thích các bước giải:
a) Ta có:
$\begin{array}{l}
A = \dfrac{{{x^2} - \sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {x + \sqrt x + 1} \right)}} - \dfrac{{x - 3}}{{\sqrt x + 1}}\left( {DK:x \ge 0} \right)\\
= \dfrac{{{x^2} - \sqrt x - \left( {x - 3} \right)\left( {x + \sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{{x^2} - \sqrt x - {x^2} - x\sqrt x + 2x + 3\sqrt x + 3}}{{\left( {\sqrt x + 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{ - x\sqrt x + 2x + 2\sqrt x + 3}}{{\left( {\sqrt x + 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{ - {{\left( {\sqrt x } \right)}^3} + 3{{\left( {\sqrt x } \right)}^2} - {{\left( {\sqrt x } \right)}^2} + 3\sqrt x - \sqrt x + 3}}{{\left( {\sqrt x + 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{\left( {\sqrt x - 3} \right)\left( { - x - \sqrt x - 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {x + \sqrt x + 1} \right)}}\\
= \dfrac{{ - \sqrt x + 3}}{{\sqrt x + 1}}
\end{array}$
Vậy $A = \dfrac{{ - \sqrt x + 3}}{{\sqrt x + 1}}$ với $x\ge 0$
b) Ta có:
$A = \dfrac{{ - \sqrt x + 3}}{{\sqrt x + 1}}$ với $x\ge 0$
$A = \dfrac{{ - \sqrt x + 3}}{{\sqrt x + 1}} = \dfrac{{ - \left( {\sqrt x + 1} \right) + 4}}{{\sqrt x + 1}} = - 1 + \dfrac{4}{{\sqrt x + 1}}$
Mà ta có:
$\begin{array}{l}
\sqrt x \ge 0,\forall x \ge 0\\
\Rightarrow \sqrt x + 1 \ge 1\\
\Rightarrow \dfrac{4}{{\sqrt x + 1}} \le 4\\
\Rightarrow - 1 + \dfrac{4}{{\sqrt x + 1}} \le 3\\
\Rightarrow A \le 3\\
\Rightarrow MaxA = 3
\end{array}$
Dấu bằng xảy ra
$\begin{array}{l}
\Leftrightarrow \sqrt x = 0\\
\Leftrightarrow x = 0
\end{array}$
Vậy $MaxA = 3 \Leftrightarrow x = 0$
