1: \(A=\dfrac{2}{\sqrt{5}+2}-\dfrac{5+\sqrt{5}}{\sqrt{5}+1}+\sqrt{14-3\sqrt{20}}\)
\(=\dfrac{2\left(\sqrt{5}-2\right)}{5-4}-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}+\sqrt{14-6\sqrt{5}}\)
\(=2\left(\sqrt{5}-2\right)-\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}\)
\(=2\sqrt{5}-4-\sqrt{5}+3-\sqrt{5}=-1\)
2:
a: \(B=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+1}{x-4\sqrt{x}+4}\)
\(=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)
\(=\dfrac{1+\sqrt{x}}{\left(\sqrt{x}-2\right)\cdot\sqrt{x}}\cdot\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
b: Để B là số tự nhiên thì \(\left\{{}\begin{matrix}B>=0\\\sqrt{x}-2⋮\sqrt{x}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\sqrt{x}-2>0\\-2⋮\sqrt{x}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>4\\\sqrt{x}\in\left\{1;2\right\}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>4\\x\in\left\{1;4\right\}\end{matrix}\right.\)
=>\(x\in\varnothing\)
