Đáp án:
`a)` `2`
`b)` $-\sqrt[3]{a}-1$ với `a\ne -1`
Giải thích các bước giải:
`a)` $(\sqrt[3]{\sqrt{9+4\sqrt{5}}}+\sqrt[3]{2+\sqrt{5}}).\sqrt[3]{\sqrt{5}-2}$
$=(\sqrt[3]{\sqrt{2^2+2.2.\sqrt{5}+5}}+\sqrt[3]{2+\sqrt{5}}).\sqrt[3]{\sqrt{5}-2}$
$=(\sqrt[3]{\sqrt{(2+\sqrt{5})^2}}+\sqrt[3]{2+\sqrt{5}}).\sqrt[3]{\sqrt{5}+2}$
$=(\sqrt[3]{|2+\sqrt{5}|}+\sqrt[3]{2+\sqrt{5}}).\sqrt[3]{\sqrt{5}-2}$
$=2\sqrt[3]{2+\sqrt{5}}.\sqrt[3]{\sqrt{5}-2}$
$=2\sqrt[3]{(\sqrt{5}+2).(\sqrt{5}-2)}$
$=2\sqrt[3]{5-2^2}=2\sqrt[3]{1}=2$
$\\$
Sửa đề câu `b` ở mẫu nhé
`b)` $\dfrac{a+\sqrt{2+\sqrt{5}}.\sqrt{\sqrt{9-4\sqrt{5}}}}{\sqrt[3]{2-\sqrt{5}}.\sqrt[3]{\sqrt{9+4\sqrt{5}}}-\sqrt[3]{a^2}+\sqrt[3]{a}}$
$=\dfrac{a+\sqrt{2+\sqrt{5}}.\sqrt{\sqrt{5-2.\sqrt{5}.2+2^2}}}{\sqrt[3]{2-\sqrt{5}}.\sqrt[3]{\sqrt{5+2.\sqrt{5}.2+2^2}}-\sqrt[3]{a^2}+\sqrt[3]{a}}$
$=\dfrac{a+\sqrt{2+\sqrt{5}}.\sqrt{\sqrt{(\sqrt{5}-2)^2}}}{\sqrt[3]{2-\sqrt{5}}.\sqrt[3]{\sqrt{(\sqrt{5}+2)^2}}-\sqrt[3]{a^2}+\sqrt[3]{a}}$
$=\dfrac{a+\sqrt{2+\sqrt{5}}.\sqrt{|\sqrt{5}-2|}}{\sqrt[3]{2-\sqrt{5}}.\sqrt[3]{|\sqrt{5}+2|}-\sqrt[3]{a^2}+\sqrt[3]{a}}$
$=\dfrac{a+\sqrt{(2+\sqrt{5}).(\sqrt{5}-2)}}{\sqrt[3]{(2-\sqrt{5}).(\sqrt{5}+2)}-\sqrt[3]{a^2}+\sqrt[3]{a}}$
$=\dfrac{a+\sqrt{5-2^2}}{\sqrt[3]{2^2-5}-\sqrt[3]{a^2}+\sqrt[3]{a}}$
$=\dfrac{a+1}{-1-\sqrt[3]{a^2}+\sqrt[3]{a}}$
$=\dfrac{(a+1).(\sqrt[3]{a}-1)}{-(\sqrt[3]{a^2}-\sqrt[3]{a}+1).(\sqrt[3]{a}+1)}$
$=\dfrac{-(a+1).(\sqrt[3]{a}+1)}{(\sqrt[3]{a^2}-\sqrt[3]{a}+1).(\sqrt[3]{a}+1)}$
$=\dfrac{-(a+1).(\sqrt[3]{a}+1)}{(\sqrt[3]{a})^3+1^3}$
$=\dfrac{-(a+1).(\sqrt[3]{a}+1)}{a+1}=-\sqrt[3]{a}-1$ `\qquad (a\ne -1)`
