phải bt có bao nhiêu số 6 mới giải đc chứ

đừng chửi mik nha, mik ms hk lp 7 àk

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

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

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

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

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

\(\Rightarrow D< 2\) (đpcm)

\(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}=\frac{3}{2}\sqrt{6}+\frac{2.1}{3}\sqrt{2.3}-\frac{4.1}{2}\sqrt{3.2}\)

\(=\frac{3}{2}\sqrt{6}+\frac{2}{3}\sqrt{6}-2\sqrt{6}=\sqrt{6}\left(\frac{3}{2}+\frac{2}{3}-2\right)\)

\(=\sqrt{6}\left(\frac{9}{6}+\frac{4}{6}-\frac{12}{6}\right)=\sqrt{6}.\frac{1}{6}=\frac{\sqrt{6}}{6}\)

Vậy \(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{6}\)

Ta có \(\sqrt[4]{49+20\sqrt{6}}=\sqrt[4]{25+10\sqrt{24}+24}=\sqrt[4]{\left(5+2\sqrt{6}\right)^2}\)

                               \(=\sqrt[4]{\left(\sqrt{3}+\sqrt{2}\right)^4}=\sqrt{3}+\sqrt{2}\)

Tương tự : \(\sqrt[4]{49-20\sqrt{6}}=\sqrt{3}-\sqrt{2}\) ( Do \(\sqrt{3}>\sqrt{2}\) )

Suy ra \(\sqrt[4]{49+20\sqrt{6}}+\sqrt[4]{49-20\sqrt{6}}=2\sqrt{3}\)