Amplificam prima fractie cu conjugata numitorului:
[tex]\it{{\dfrac{ ^{\sqrt7+\sqrt2)}1}{\sqrt7-\sqrt2} =\dfrac{\sqrt7+\sqrt2}{7-2}=\dfrac{\sqrt7+\sqrt2}{5}}[/tex]
Analog se obtine :
[tex]\it{{\dfrac{1}{\sqrt7+\sqrt2} =\dfrac{\sqrt7-\sqrt2}{7-2}=\dfrac{\sqrt7-\sqrt2}{5}}[/tex]
[tex]\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{\sqrt7+\sqrt2+\sqrt7-\sqrt2}{5} = \dfrac{2\sqrt7}{5}=\dfrac{\sqrt{28}}{5}[/tex]
[tex]16\ \textless \ 28\ \textless \ 36 \Rightarrow \sqrt{16}\ \textless \ \sqrt{28}\ \textless \ \sqrt{36} \Rightarrow 4\ \textless \ \sqrt{28}\ \textless \ 6|_{:5}[/tex]
[tex]\Longrightarrow \dfrac{4}{5}\ \textless \ \dfrac{\sqrt{28}}{5}\ \textless \ \dfrac{6}{5}[/tex]
Deci, [tex]\dfrac{1}{a}+\dfrac{1}{b} \in \left(\dfrac{4}{5},\ \dfrac{6}{5} \right)[/tex]
