Question

$Calculati~partea~intreaga~a~numarului~~ \sqrt[3]{24+ \sqrt[3]{24+... \sqrt[3]{24} } } .$

Answer 1

$\displaystyle Era~bine~daca~spuneai~si~cati~termeni~apar~sub~radical~(probabil~n). \\ \\ In~fine,~notez~numarul~cu~N. \\ \\ Avem~N\ \textgreater \ \sqrt[3]{24}\ \textgreater \ \sqrt[3]{8}=2. \\ \\ Folosind~faptul~ca~24\ \textless \ 27,~obtinem: \\ \\ N\ \textless \ \sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{...+\sqrt[3]{24+\sqrt[3]{27}}}}}= \\ \\ =\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{...+\sqrt[3]{24+3}}}}= \\ \\ =\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{...+\sqrt[3]{27}}}}= \\ \\ =...=\sqrt[3]{24+\sqrt[3]{27}}=\sqrt[3]{24+3}=3. \\ \\ Deci~N \in (2,3).~Rezulta~[N]=2.$