\(\dfrac{1}{\sqrt{5}+\sqrt{3}}-\dfrac{1}{\sqrt{5}-\sqrt{3}}\\ =\dfrac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\dfrac{\sqrt{5}+\sqrt{3}}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}\\ =\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{5-3}\\ =\dfrac{-2\sqrt{3}}{2}\\ =-\sqrt{3}\)
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\(\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}\\ =\dfrac{\sqrt{\left(\sqrt{3}\right)^2-2\cdot\sqrt{3}\cdot1+1^2}}{\sqrt{6}-\sqrt{2}}\\ =\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}\\ =\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}\\ =\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
\(\dfrac{1}{\sqrt{5}+\sqrt{3}}-\dfrac{1}{\sqrt{5}-\sqrt{3}}\\ =\dfrac{\sqrt{5}-\sqrt{3}}{5-3}-\dfrac{\sqrt{5}+\sqrt{3}}{5-3}\\ =\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{2}\\ =\dfrac{-2\sqrt{3}}{2}=-\sqrt{3}\\ ---\\ \dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}\\ =\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\left|\sqrt{3}-1\right|}{\sqrt{2}\left(\sqrt{3}-1\right)}\\ =\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
\(---\\ \sqrt{127-48\sqrt{7}}-\sqrt{127+48\sqrt{7}}\\ =\sqrt{\left(3\sqrt{7}\right)^2-2.3\sqrt{7}.8+8^2}-\sqrt{\left(3\sqrt{7}\right)^2+2.3\sqrt{7}.8+8^2}\\ =\sqrt{\left(3\sqrt{7}-8\right)^2}-\sqrt{\left(3\sqrt{7}+8\right)^2}\\ =\left|3\sqrt{7}-8\right|-\left|3\sqrt{7}+8\right|\\ =8-3\sqrt{7}-3\sqrt{7}-8\\ =-6\sqrt{7}\\ ---\\ \dfrac{\sqrt{10}+\sqrt{6}}{2\sqrt{5}+\sqrt{12}}=\dfrac{\sqrt{10}+\sqrt{6}}{2\sqrt{5}+2\sqrt{3}}\\ =\dfrac{\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)}{2\left(\sqrt{5}+\sqrt{3}\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)
#$\mathtt{Toru}$
\(\sqrt{127-48\sqrt{7}}-\sqrt{127+48\sqrt{7}}\\ =\sqrt{8^2-2\cdot8\cdot3\sqrt{7}+\left(3\sqrt{7}\right)^2}-\sqrt{8^2+2\cdot8\cdot3\sqrt{7}+\left(\sqrt{7}\right)^2}\\ =\sqrt{\left(8-3\sqrt{7}\right)^2}-\sqrt{\left(8+3\sqrt{7}\right)^2}\\ =8-3\sqrt{7}-8-3\sqrt{7}\\ =-6\sqrt{7}\)
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\(\dfrac{\sqrt{10}+\sqrt{6}}{2\sqrt{5}+\sqrt{12}}\\ =\dfrac{\sqrt{10}+\sqrt{6}}{2\sqrt{5}+2\sqrt{3}}\\ =\dfrac{\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)}{2\left(\sqrt{5}+\sqrt{3}\right)}\\ =\dfrac{\sqrt{2}}{2}\)
