$C=\sqrt{8+2\sqrt{15}}-\sqrt{5-2\sqrt{6}}\\=\sqrt{8+2.\sqrt{5}.\sqrt{3}}-\sqrt{5-2.\sqrt{2}.\sqrt{3}}\\=\sqrt{(\sqrt{5}+\sqrt{3})^2}-\sqrt{(\sqrt{3}-\sqrt{2})^2}\\=|\sqrt{5}+\sqrt{3}|-|\sqrt{3}-\sqrt{2}|\\=\sqrt{5}+\sqrt{3}-(\sqrt{3}-\sqrt{2}) \text{ (vì }\sqrt{3}>\sqrt{2})\\=\sqrt{5}+\sqrt{2}$
$H=\sqrt{19-3\sqrt{40}}-\sqrt{19+3\sqrt{40}}\\=\sqrt{19-3.2.\sqrt{10}}-\sqrt{19+3.2.\sqrt{10}}\\=\sqrt{(\sqrt{10}-3)^2}-\sqrt{(\sqrt{10}+3)^2}\\=|\sqrt{10}-3|-|\sqrt{10}+3|\\=\sqrt{10}-3-(\sqrt{10}+3)\text{ (vì }\sqrt{10}>3)\\=-6$
$G=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\\=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2.1.\sqrt{3}}}}\\=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{(\sqrt{3}+1)^2}}}\\=\sqrt{6+2\sqrt{2}\sqrt{3-|\sqrt{3}+1|}}\\=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\\=\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\\=\sqrt{6+2\sqrt{2(2-\sqrt{3})}}\\=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\\=\sqrt{6+2\sqrt{4-2.1.\sqrt{3}}}\\=\sqrt{6+2\sqrt{(\sqrt{3}-1)^2}}\\=\sqrt{6+2|\sqrt{3}-1|}\\=\sqrt{6+2(\sqrt{3}-1)}\text{ (vì }\sqrt{3}>1)\\=\sqrt{6+2\sqrt{3}-2}\\=\sqrt{4+2\sqrt{3}}\\=\sqrt{4+2.1.\sqrt{3}}\\=\sqrt{(\sqrt{3}+1)^2}\\=|\sqrt{3}+1|\\=\sqrt{3}+1$
$D=\sqrt{28-10\sqrt{3}}+\sqrt{4+2\sqrt{3}}\\=\sqrt{28-2.5.\sqrt{3}}+\sqrt{4+2.1.\sqrt{3}}\\=\sqrt{(5-\sqrt{3})^2}+\sqrt{(\sqrt{3}+1)^2}\\=|5-\sqrt{3}|+|\sqrt{3}+1|\\=5-\sqrt{3}+\sqrt{3}+1\text{ (vì }5>\sqrt{3})\\=6$
$E=\sqrt{14-6\sqrt{5}}-\sqrt{21-8\sqrt{5}}\\=\sqrt{14-2.3.\sqrt{5}}-\sqrt{21-2.4.\sqrt{5}}\\=\sqrt{(3-\sqrt{5})^2}-\sqrt{(4-\sqrt{5})^2}\\=|3-\sqrt{5}|-|4-\sqrt{5}|\\=3-\sqrt{5}-(4-\sqrt{5})\text{ (vì }3>\sqrt{5},\,4>\sqrt{5})\\=-1$
$F=\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}\\=\sqrt{21-2.3\sqrt{2}.\sqrt{3}}+\sqrt{9+2.\sqrt{3}.\sqrt{6}}-2\sqrt{\dfrac{2.(6+3\sqrt{3})}{2}}\\=\sqrt{(3\sqrt{2}-\sqrt{3})^2}+\sqrt{(\sqrt{6}+\sqrt{3})^2}-2.\dfrac{\sqrt{12+6\sqrt{3}}}{\sqrt{2}}\\=|3\sqrt{2}-\sqrt{3}|+|\sqrt{6}+\sqrt{3}|-\sqrt{2}.\sqrt{12+2.3.\sqrt{3}}\\=3\sqrt{2}-\sqrt{3}+\sqrt{6}+\sqrt{3}-\sqrt{2}.\sqrt{(3+\sqrt{3})^2}\\=3\sqrt{2}+\sqrt{6}-\sqrt{2}.|3+\sqrt{3}|\\=3\sqrt{2}+\sqrt{6}-\sqrt{2}.(3+\sqrt{3})\\=3\sqrt{2}+\sqrt{6}-3\sqrt{2}-\sqrt{6}\\=0$
$I=\sqrt{4+\sqrt{15}}-\sqrt{7-3\sqrt{5}}\\=\sqrt{\dfrac{2.(4+\sqrt{15})}{2}}-\sqrt{\dfrac{2.(7-3\sqrt{5})}{2}}\\=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}-\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{2}}\\=\dfrac{\sqrt{8+2.\sqrt{5}.\sqrt{3}}-\sqrt{14-2.3.\sqrt{5}}}{\sqrt{2}}\\=\dfrac{\sqrt{(\sqrt{5}+\sqrt{3})^2}-\sqrt{(3-\sqrt{5})^2}}{\sqrt{2}}\\=\dfrac{|\sqrt{5}+\sqrt{3}|-|3-\sqrt{5}|}{\sqrt{2}}\\=\dfrac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}\\=\dfrac{2\sqrt{5}-3+\sqrt{3}}{\sqrt{2}}$
$J=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\\=\sqrt{\dfrac{2.(2+\sqrt{3})}{2}}+\sqrt{\dfrac{2.(2-\sqrt{3}}{2}}\\=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}+\dfrac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\\=\dfrac{\sqrt{4+2.1.\sqrt{3}}+\sqrt{4-2.1.\sqrt{3}}}{\sqrt{2}}\\=\dfrac{\sqrt{(\sqrt{3}+1)^2}+\sqrt{(\sqrt{3}-1)^2}}{\sqrt{2}}\\=\dfrac{|\sqrt{3}+1|+|\sqrt{3}-1|}{\sqrt{2}}\\=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}\\=\dfrac{2\sqrt{3}}{\sqrt{2}}\\=\sqrt{6}$
