Question

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

Tính P=\(\frac{x\cdot y}{x+y}\)

Answer 1

Đặt \(t=\sqrt[3]{2}\) và  

=> \(x=\frac{t^3}{2t+t^3+t^2}=\frac{t^2}{t^2+t+2}\)

=> \(y=\frac{3t^3}{2t-t^3+t^2}=\frac{3t^2}{t-t^2+2}\)

\(xy=\frac{3t^4}{\left(t+2+t^2\right)\left(t+2-t^2\right)}\)

\(x+y=\frac{t^2}{t^2+t+2}+\frac{3t^2}{t-t^2+2}=\frac{t^3-t^4+2t^2+3t^4+3t^3+6t^2}{\left(t^2+t+2\right)\left(t-t^2+2\right)}=\frac{2t^2\left(t^2+2t+4\right)}{\left(t^2+t+2\right)\left(t-t^2+2\right)}\)

Suy ra : \(\frac{xy}{x+y}=\frac{3t^4}{\left(t^2+t+2\right)\left(t+2-t^2\right)}:\frac{2t^2\left(t^2+2t+4\right)}{\left(t^2+t+2\right)\left(t-t^2+2\right)}\)

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