Question

Chứng minh rằng biểu thức A = \(A=\sqrt[3]{70-\sqrt{4901}}+\sqrt[3]{70+\sqrt{4901}}\)là số nguyên.

Answer 1

\(A=\sqrt[3]{70-\sqrt{4901}}+\sqrt[3]{70+\sqrt{4901}}\)

\(A^3=70-\sqrt{4901}+3\left(\sqrt[3]{70-\sqrt{4901}}\right)^2\sqrt{70+\sqrt{4901}}+3\left(\sqrt[3]{70-\sqrt{4901}}\right)\left(\sqrt[3]{70+\sqrt{4901}}\right)^2+70+\sqrt{4901}\) =\(140-3\left(\sqrt[3]{70-4901\left(7+\sqrt{4901}\right)}\right)\left(\dfrac{\sqrt[3]{70-\sqrt{4901}}+\sqrt[3]{70+4901}}{A}\right)\)\(A^3=140-3D\Rightarrow A^3-3A-140=0\Rightarrow A=5\)\(\Rightarrowđpcm\)