Đáp án:
$\dfrac{1}{2+\sqrt[3]{2}+\sqrt[3]{4}}=1-\dfrac{\sqrt[3]{4}}{2}$
Giải thích các bước giải:
Ta có:
$\quad \dfrac{1}{2+\sqrt[3]{2}+\sqrt[3]{4}}$
$= \dfrac{1}{\sqrt[3]{2^3}+\sqrt[3]{4}+\sqrt[3]{2}}$
$= \dfrac{1}{\sqrt[3]{2}.(\sqrt[3]{2^2}+\sqrt[3]{2}+1)}$
$=\dfrac{\sqrt[3]{2}-1}{\sqrt[3]{2}.(\sqrt[3]{2}-1).[(\sqrt[3]{2})^2+\sqrt[3]{2}.1+1^2]}$
$=\dfrac{\sqrt[3]{2}-1}{\sqrt[3]{2}.[(\sqrt[3]{2})^3-1^3]}$
$=\dfrac{\sqrt[3]{2}-1}{\sqrt[3]{2}.1}$
$=\dfrac{\sqrt[3]{2}}{\sqrt[3]{2}}-\dfrac{1}{\sqrt[3]{2}}$
$=1-\dfrac{\sqrt[3]{2^2}}{\sqrt[3]{2^2}.\sqrt[3]{2}}$
$=1-\dfrac{\sqrt[3]{4}}{\sqrt[3]{2^3}}=1-\dfrac{\sqrt[3]{4}}{2}$
Vậy: $\dfrac{1}{2+\sqrt[3]{2}+\sqrt[3]{4}}=1-\dfrac{\sqrt[3]{4}}{2}$
