Giải thích các bước giải:

a.Ta có:

$\begin{aligned}
A &= \sqrt{\dfrac{x\sqrt{x} + 1}{\sqrt{x} + 1} - \sqrt{x}} \\
&= \sqrt{\dfrac{(\sqrt{x})^3 + 1^3}{\sqrt{x} + 1} - \sqrt{x}} \\
&= \sqrt{\dfrac{(\sqrt{x} + 1)(x - \sqrt{x} + 1)}{\sqrt{x} + 1} - \sqrt{x}} \\
&= \sqrt{(x - \sqrt{x} + 1) - \sqrt{x}} \\
&= \sqrt{x - 2\sqrt{x} + 1} \\
&= \sqrt{(\sqrt{x} - 1)^2} \\
&= |\sqrt{x} - 1|
\end{aligned}$

b.Ta có:

 $\begin{aligned}
x &= 6 - 2\sqrt{5} \\
&= 5 - 2\sqrt{5} \cdot 1 + 1 \\
&= (\sqrt{5})^2 - 2\sqrt{5} + 1^2 \\
&= (\sqrt{5} - 1)^2
\end{aligned}$

$\begin{aligned}
\Rightarrow A &= |\sqrt{(\sqrt{5} - 1)^2} - 1| \\
&= |(\sqrt{5} - 1) - 1| \quad (\text{vì } \sqrt{5} > 1 \text{ nên } \sqrt{5} - 1 > 0) \\
&= |\sqrt{5} - 2|
\end{aligned}$

$\Rightarrow A = \sqrt{5} - 2$ vì $\sqrt5-2>0$