Đáp án:
$D=1$
Giải thích các bước giải:
$D=\sqrt[3]{2+\sqrt{5}}-\sqrt[3]{\sqrt{5}-2}$
$=\dfrac{1}{2}.(2\sqrt[3]{2+\sqrt{5}}-2\sqrt[3]{\sqrt{5}-2})$
$=\dfrac{1}{2}.(\sqrt[3]{16+8\sqrt{5}}+\sqrt[3]{16-8\sqrt{5}})$
$=\dfrac{1}{2}.(\sqrt[3]{1+3\sqrt{5}+15+5\sqrt{5}}+\sqrt[3]{1-3\sqrt{5}+15-5\sqrt{5}})$
$=\dfrac{1}{2}.(\sqrt[3]{(1+\sqrt{5})^3}+\sqrt[3]{(1-\sqrt{5})^3})$
$=\dfrac{1}{2}.(1+\sqrt{5}+1-\sqrt{5})$
$=\dfrac{1}{2}.2$
$=1$
