ta có:\(\sqrt{100+1}+\sqrt{100-1}< 2\sqrt{100}=20\)
=>\(\frac{\sqrt{101}+\sqrt{99}}{2}< 10\)
<=>\(\frac{\sqrt{101} +\sqrt{99}}{\left(\sqrt{101}+\sqrt{99}\right)\left(\sqrt{101}-\sqrt{99}\right)}\) <10
<=>\(\frac{1}{\sqrt{101}-\sqrt{99}}\) <10
=>\(\sqrt{100}-\sqrt{99}>\frac{1}{10}=0,1\) (đpcm)