Question

Chứng minh rằng :

D=\(\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6++...+\sqrt[3]{6}}}}< 2\)

Answer 1

\(D=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6+\sqrt[3]{6}}}}}\)

\(\Rightarrow D< \sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6+\sqrt[3]{8}}}}}\)

\(\Rightarrow D< \sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6+2}}}}\)

\(\Rightarrow D< \sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{8}}}}\)

\(\Rightarrow D< \sqrt[3]{6+\sqrt[3]{8}}=\sqrt[3]{6+2}=\sqrt[3]{8}\)

\(\Rightarrow D< 2\) (đpcm)