Question

cho x = \(\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}\)  ;  y = \(\frac{6}{2\sqrt[3]{2}-2+\sqrt[3]{4}}\)

Tính P = \(\frac{xy}{x+y}\)

giúp mình nha. mình trục căn thức ở mẫu từng cái rối quá. khó làm . 

Answer 1

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

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

\(=\frac{6\left(\sqrt[3]{4}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{4}\right)^3+\left(\sqrt[3]{2}\right)^3}=\sqrt[3]{4}+\sqrt[3]{2}\)

\(P=\frac{xy}{x+y}=\frac{\sqrt[3]{4^2}-\sqrt[3]{2^2}}{2\sqrt[3]{4}}=\frac{\sqrt[3]{4}-1}{2}\)