a.
$\sqrt{\dfrac{1}{5-2\sqrt{6}}}=\sqrt{\dfrac{1}{(\sqrt{3}-\sqrt{2})^2}}=\dfrac{1}{\sqrt{3}-\sqrt{2}}=\sqrt{3}+\sqrt{2}\\5+2\sqrt{6}=(\sqrt{3}+\sqrt{2})^2\\\sqrt{3}+\sqrt{2}>1\\\Rightarrow 5+2\sqrt{6}>\sqrt{\dfrac{1}{5-2\sqrt{6}}}$
b.
$A=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\\\Rightarrow \sqrt{2}A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\\=\sqrt{(\sqrt{7}+1)^2}-\sqrt{(\sqrt{7}-1)^2}-2\\A=0<1$
c.
$\dfrac{1}{\sqrt{30}-\sqrt{29}}=\sqrt{30}+\sqrt{29}\\\dfrac{1}{\sqrt{29}-\sqrt{28}}=\sqrt{29}+\sqrt{28}\\\Rightarrow \sqrt{30}-\sqrt{29}<\sqrt{29}-\sqrt{28}$
d.
$(\sqrt{1999}+\sqrt{2001})^2 <2(1999+2001)=8000\\\Rightarrow \sqrt{1999}+\sqrt{2001} < 2\sqrt{2000}$