Question

[Như]

trục căn thức ở mẫu:

\(A=\frac{2}{2\cdot\sqrt[3]{2}+2+\sqrt[3]{4}}\)

\(B=\frac{6}{2\cdot\sqrt[3]{3}-2+\sqrt[3]{4}}\)

Answer 1

[ngonhuminh]

\(A=\dfrac{2}{2.\sqrt[3]{2}+2+\sqrt[3]{2^2}}=\dfrac{2}{\left(\sqrt[3]{2}\right)^2+2.\left(\sqrt[3]{2}\right)+\left(\sqrt{2}\right)^2}\)

\(A=\dfrac{2.\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)}{\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)\left[\left(\sqrt[3]{2}\right)^2+2.\left(\sqrt[3]{2}\right)+\left(\sqrt{2}\right)^2\right]}=\dfrac{2.\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)}{\left(\sqrt[3]{2}\right)^3-\left(\sqrt{2}\right)^3}=\dfrac{2.\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)}{2-2\sqrt{2}}\)

\(A=\dfrac{2\left[.\left(\sqrt[3]{2}\right)-\left(\sqrt{2}\right)\right].\left(1+\sqrt{2}\right)}{2\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}=\left(\sqrt{2}+1\right)\left(\sqrt{2}-\sqrt[3]{2}\right)\)

Answer 2

[Ngọc Hiền]

ở phân thức A nhân cả tử và mẫu cho: (2\(\sqrt[3]{2}\))²-2.\(\sqrt[3]{2}\left(2+\sqrt[3]{4}\right)+\left(2-\sqrt[3]{4}\right)^2\)

ở phân thức B nhân cả tử và mẫu cho :(2\(\sqrt[3]{2}\))²+\(2.\sqrt[3]{2}\left(2-\sqrt[3]{4}\right)+\left(2-\sqrt[3]{4}\right)^2\)