Question

Cho \(b=\sqrt[3]{2020}\). Tính:

\(Q=\sqrt[3]{\dfrac{b^3-3b+\left(b^2-1\right)\sqrt{b^2-4}}{2}}+\sqrt[3]{\dfrac{b^3-3b-\left(b^2-1\right)\sqrt{b^2-4}}{2}}\)

Answer 1

\(Q^3=\dfrac{b^3-3b+\left(b^2-1\right)\sqrt{b^2-4}+b^3-3b-\left(b^2-1\right)\sqrt{b^2-4}}{2}+3Q\sqrt[3]{\dfrac{\left(b^3-3b+\left(b^2-1\right)\sqrt{b^2-4}\right)\left(b^3-3b-\left(b^2-1\right)\sqrt{b^2-4}\right)}{4}}\)

\(Q^3=\dfrac{2b^3-6b}{2}+3Q\sqrt[3]{\dfrac{\left(b^3-3b\right)^2-\left(b^2-1\right)^2\left(b^2-4\right)}{4}}\\ Q^3=b^3-3b+3Q\sqrt[3]{\dfrac{b^6-6b^4+9b^2-b^6+6b^4-9b^2+4}{4}}\\ Q^3=b^3-3b+3Q\sqrt[3]{\dfrac{4}{4}}=b^3-3b+3Q\\ \Leftrightarrow Q^3-3Q=b^3-3b\\ \Leftrightarrow Q\left(Q^2-3\right)=b\left(b^2-3\right)\)

\(\Leftrightarrow Q=b=\sqrt[3]{2020}\) (hmm ko chắc)