\(\sqrt{19}+\sqrt{21}=\sqrt{\left(\sqrt{19}+\sqrt{21}\right)^2}=\sqrt{40+2\sqrt{19\cdot21}}=\sqrt{40+2\sqrt{\left(20-1\right)\left(20+1\right)}}=\sqrt{40+2\sqrt{20^2-1}}< \sqrt{40+2\sqrt{20^2}}=\sqrt{80}=2\sqrt{20}\)

\(a,\left(\sqrt{2}+\sqrt{11}\right)^2=12+2\sqrt{22}\\ \left(\sqrt{3}+5\right)^2=28+10\sqrt{3}\)

Ta thấy \(12< 28;2\sqrt{22}=\sqrt{88}< \sqrt{300}=10\sqrt{3}\)

Nên \(\sqrt{2}+\sqrt{11}< \sqrt{3}+5\)

\(b,\left(\sqrt{21}-\sqrt{5}\right)^2=26-2\sqrt{105}\\ \left(\sqrt{20}-\sqrt{6}\right)^2=26-2\sqrt{120}\)

Vì \(\sqrt{105}< \sqrt{120}\Rightarrow-2\sqrt{105}>-2\sqrt{120}\)

Nên \(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)

\(\frac{\sqrt{21} - \sqrt{13}}{35 - 2\sqrt{273}} = \frac{\sqrt{21} - \sqrt{13}}{(\sqrt{21} - \sqrt{13})^2} = \frac{1}{\sqrt{21} - \sqrt{13}}\)

\(=\frac{\sqrt{21}+\sqrt{13}}{8}>\frac{\sqrt{16}+\sqrt9}{8}=\frac{4+3}{8}=\frac78\)

\(\frac{\sqrt{10} - \sqrt{5}}{16 - 10\sqrt{2}}=\frac{\sqrt{5}(\sqrt{2} - 1)}{16 - 10\sqrt{2}}=\frac{\sqrt{5}}{6\sqrt{2} - 4}=\frac{\sqrt5}{2\left(3\sqrt2-2\right)}\)

\(\frac{\sqrt5}{2\left(3\sqrt2-2\right)}-\frac18=\frac{8\sqrt5-6\sqrt2+4}{16\left(3\sqrt2-2\right)}=\frac{\sqrt{320}-\sqrt{72}+4}{16\left(3\sqrt2-2\right)}>\frac{17-8+4}{16\left(3\sqrt2-2\right)}=\frac{13}{16\left(3\sqrt2-2\right)}>5>\frac18\)

Do đó: \(\frac{\sqrt{21}-\sqrt{13}}{35-2\sqrt{273}}+\frac{\sqrt{10}-\sqrt5}{16-10\sqrt2}\) >1(ĐPCM)