Question

a= √7-√2
b= √7+√2

1/a + 1/b inclus in intervalul (4/5; 6/5)

Answer 1

Deci
1)formam suma noastra, adica inlocuim in formula ta (1/a+1/b)
obtinem 
(1/√7-√2) + (1/√7+√2)
2) efectuind calculele simple obtinem (2√7)/7-4
3)introducem sub radical si obtinem √28/3 
prin aproximare √28=5.3 iar 5.3/3 este 1.7(6) 
rezulta ca expresia data de tine nu apartine intervalului dat

seara buna

Answer 2

Amplificam prima fractie cu conjugata numitorului:

$\it{{\dfrac{ ^{\sqrt7+\sqrt2)}1}{\sqrt7-\sqrt2} =\dfrac{\sqrt7+\sqrt2}{7-2}=\dfrac{\sqrt7+\sqrt2}{5}}$

Analog se obtine :

$\it{{\dfrac{1}{\sqrt7+\sqrt2} =\dfrac{\sqrt7-\sqrt2}{7-2}=\dfrac{\sqrt7-\sqrt2}{5}}$

$\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{\sqrt7+\sqrt2+\sqrt7-\sqrt2}{5} = \dfrac{2\sqrt7}{5}=\dfrac{\sqrt{28}}{5}$

$16\ \textless \ 28\ \textless \ 36 \Rightarrow \sqrt{16}\ \textless \ \sqrt{28}\ \textless \ \sqrt{36} \Rightarrow 4\ \textless \ \sqrt{28}\ \textless \ 6|_{:5}$

$\Longrightarrow \dfrac{4}{5}\ \textless \ \dfrac{\sqrt{28}}{5}\ \textless \ \dfrac{6}{5}$

Deci, $\dfrac{1}{a}+\dfrac{1}{b} \in \left(\dfrac{4}{5},\ \dfrac{6}{5} \right)$