[tex]a\ ?\ b\\
\sqrt{99}+\sqrt{101}\ \ \sqrt{98}+\sqrt{102}\\
\sqrt{99}-\sqrt{98}\ \ \sqrt{102}-\sqrt{101}\\
\text{Daca rationalizam cele doua fractii obtinem:}\\
\dfrac{99-98}{\sqrt{99}+\sqrt{98}}\ \ \dfrac{102-101}{\sqrt{102}+\sqrt{101}}\\
\dfrac{1}{\sqrt{99}+\sqrt{98}}\ \ \dfrac{1}{\sqrt{102}+\sqrt{101}}\\
\text{Deoarece } \sqrt{99}+\sqrt{98} \ \textless \ \sqrt{102}+\sqrt{101}\Rightarrow \dfrac{1}{\sqrt{99}+\sqrt{98}} \ \textgreater \ \dfrac{1}{\sqrt{102}+\sqrt{101}}\\
\text{Deci}\ a\ \textgreater \ b
[/tex]
