2) \(B=\dfrac{\sqrt{3-\sqrt{5}}}{\sqrt{2}}-\dfrac{\sqrt{5}-1}{\sqrt{5}+1}=\dfrac{\sqrt{\dfrac{6-2\sqrt{5}}{2}}}{\sqrt{2}}-\dfrac{\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=\dfrac{\sqrt{\dfrac{\left(\sqrt{5}-1\right)^2}{2}}}{\sqrt{2}}-\dfrac{6-2\sqrt{5}}{4}=\dfrac{\dfrac{\sqrt{5}-1}{\sqrt{2}}}{\sqrt{2}}-\dfrac{6-2\sqrt{5}}{4}\)
\(=\dfrac{\sqrt{5}-1}{2}-\dfrac{6-2\sqrt{5}}{4}=\dfrac{2\sqrt{5}-2-6+2\sqrt{5}}{4}=\sqrt{5}-2\)
3) \(x-3\sqrt{x}+2=0\Leftrightarrow x-\sqrt{x}-2\sqrt{x}+2=0\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
