Question

Daca a=(√7+2) la puterea-1,demostrati ca a∈(5 supra 3 ,2)

Answer 1

$a=(\sqrt7-2)^{-1} = \dfrac{1}{\sqrt7-2} =\dfrac{\sqrt7+2}{7-4} = \dfrac{\sqrt7+2}{3}$

Acum comparam numarul cu extremitatea stanga a intervalului dat.

$\dfrac{\sqrt7+2}{3} \ \textgreater \ \dfrac{5}{3} \Leftrightarrow \sqrt7+2 \ \textgreater \ 5 \Leftrightarrow \sqrt7 \ \textgreater \ 3 \ \ (Fals)$

Deci :

$a\not{\in} \left(\dfrac{5}{3},\ 2 \right)$