Question

Tìm phần nguyên của \(A=\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6}}}}+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6}}}}\)

Answer 1

\(\sqrt{6+\sqrt{6+...+\sqrt{6}}}>\sqrt{6}=\sqrt{\frac{150}{25}}>\sqrt{\frac{144}{25}}=\frac{12}{5}\)

\(\sqrt[3]{6+\sqrt[3]{6+...+\sqrt[3]{6}}}>\sqrt[3]{6}=\sqrt[3]{\frac{750}{125}}>\sqrt[3]{\frac{729}{125}}=\frac{9}{5}\)

\(\Rightarrow A>\frac{12}{5}+\frac{9}{5}=\frac{21}{5}>4\)

\(\sqrt{6+\sqrt{6+...+\sqrt{6}}}< \sqrt{6+\sqrt{6+...+\sqrt{9}}}=3\)

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

\(\Rightarrow A< 3+2=5\)

\(\Rightarrow4< A< 5\Rightarrow\left[A\right]=4\)