Question

Cho x=\(\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}};y=\frac{6}{2\sqrt[3]{2}-2+\sqrt[3]{4}}.TínhM=xy\left(x^2-y^2\right)\)

Answer 1

Ta có:\(x=\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}=\frac{2}{\left(\sqrt[3]{4}\right)^2+\sqrt[3]{4}.\sqrt[3]{2}+\left(\sqrt[3]{2}\right)^2}=\frac{2\left(\sqrt[3]{4}-\sqrt[3]{2}\right)}{\left[\left(\sqrt[3]{4}\right)^2+\sqrt[3]{4}.\sqrt[3]{2}+\left(\sqrt[3]{2}\right)^2\right]\left(\sqrt[3]{4}-\sqrt[3]{2}\right)}\)

\(=\sqrt[3]{4}-\sqrt[3]{2}\). 

Tương tự 

Answer 2

\(y=\sqrt[3]{4}+\sqrt[3]{2}\). Thay x, y vào ta tính được:

\(M=8\sqrt[3]{4}-16\sqrt[3]{2}\)