Question

Trục căn thức ở mẫu :

a) \(\frac{1}{\sqrt{13-\sqrt{48}}}\)

b) \(\frac{2}{\sqrt[3]{4}+\sqrt[3]{2}+2}\)

 

Answer 1

\(\frac{1}{\sqrt{13-\sqrt{48}}}=\frac{1}{\sqrt{12+1+2\cdot2\sqrt{3}}}=\frac{1}{2\sqrt{3}+1}=\frac{-1+2\sqrt{3}}{11}\)\

Answer 2

Câu b nè: 

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

Đặt: \(\sqrt[3]{2}=a\)

=> \(B=\frac{a^3}{a^3+a^2+a}=\frac{a^2}{a^2+a+1}=\frac{a^2\left(a-1\right)}{\left(a^2+a+1\right)\left(a-1\right)}=\frac{a^3-a^2}{a^3-1}=\frac{2-\sqrt[3]{4}}{2-1}=2-\sqrt[3]{4}\)

Vậy \(B=2-\sqrt[3]{4}\)