a) \(A=\sqrt[3]{70-\sqrt{4901}}+\sqrt[3]{70+\sqrt{4901}}\)
\(\Leftrightarrow A^3=70-\sqrt{4901}+70+\sqrt{4901}+3\sqrt[3]{\left(70-\sqrt{4901}\right)\left(70+\sqrt{4901}\right)}\cdot A\)
\(\Leftrightarrow A^3=140+3\sqrt[3]{4900-4901}\cdot A\)
\(\Leftrightarrow A^3=140-3A\)
\(\Leftrightarrow A^3+3A-140=0\)
\(\Leftrightarrow\left(A-5\right)\left(A^2+5A+28\right)=0\)
\(\Leftrightarrow A=5\) ( do \(A^2+5A+28>0\forall A\) )
Vậy \(A=5\)
\(B=\frac{1}{\sqrt[3]{6}+\sqrt[3]{4}+\sqrt[3]{9}}+\sqrt[3]{2}\\ =\frac{1}{\sqrt[3]{6}+\sqrt[3]{4}+\sqrt[3]{9}}+\frac{\sqrt[3]{2}\left(\sqrt[3]{6}+\sqrt[3]{4}+\sqrt[3]{9}\right)}{\sqrt[3]{6}+\sqrt[3]{4}+\sqrt[3]{9}}\\ =\frac{1+\sqrt[3]{12}+2+\sqrt[3]{18}}{\sqrt[3]{6}+\sqrt[3]{4}+\sqrt[3]{9}}=\frac{3+\sqrt[3]{12}+\sqrt[3]{18}}{\sqrt[3]{6}+\sqrt[3]{4}+\sqrt[3]{9}}\\ =\frac{\sqrt[3]{3}\left(\sqrt[3]{6}+\sqrt[3]{4}+\sqrt[3]{9}\right)}{\sqrt[3]{6}+\sqrt[3]{4}+\sqrt[3]{9}}=\sqrt[3]{3}\)
