Giúp em bài này với ạ , khó quá

Giải thích các bước giải:

Câu 2:

a.Ta có:

$A=2\sqrt3+\dfrac{10+2\sqrt8}2-\sqrt{12}$

$\begin{aligned}
& 2\sqrt{3} + \dfrac{10+2\sqrt{8}}{2} - \sqrt{12} \\
&= 2\sqrt{3} + \dfrac{2(5+\sqrt{8})}{2} - \sqrt{4 \cdot 3} \\
&= 2\sqrt{3} + (5 + \sqrt{8}) - 2\sqrt{3} \\
&= (2\sqrt{3} - 2\sqrt{3}) + 5 + \sqrt{4 \cdot 2} \\
&= 0 + 5 + 2\sqrt{2} \\
&= 5 + 2\sqrt{2}
\end{aligned}$

b.Ta có:

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

$\to B = \left( \dfrac{1}{\sqrt{x}-3} - \dfrac{1}{\sqrt{x}} \right) \cdot \dfrac{x-3\sqrt{x}}{2025}$

$\to \begin{aligned}
B &= \dfrac{3}{\sqrt{x}(\sqrt{x}-3)} \cdot \dfrac{\sqrt{x}(\sqrt{x}-3)}{2025} \\
&= \dfrac{3}{2025} \\
&= \dfrac{1}{675}
\end{aligned}$