Question

\(\dfrac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\)

Answer 1

\(\dfrac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\)

=\(\dfrac{1}{\sqrt[3]{4-\sqrt{15}}}+\dfrac{\sqrt[3]{4-\sqrt{15}}}{1}\)

=\(\dfrac{\left(\sqrt[3]{4-\sqrt{15}}\right)^2}{\sqrt[3]{4-\sqrt{15}}}\)

=\(\sqrt[3]{4-\sqrt{15}}\).

Answer 2

mình nhầm từ bước 3 thực xin lỗi:

=\(\dfrac{1+\sqrt[3]{\left(4-\sqrt{15}\right)^2}}{\sqrt[3]{4-\sqrt{15}}}\)

=\(=\dfrac{1+\sqrt[3]{31-8\sqrt{15}}}{\sqrt[3]{4-\sqrt{15}}}\)