Question

transforme os radicais duplos em soma/diferença de radicais simples abaixo
$\sqrt{3 + \sqrt{8} }$
$\sqrt{11 - \sqrt{21} }$
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Answer 1

Utilize a fórmula:

$\sqrt{A\pm \sqrt{B}}=\sqrt{\dfrac{A+\sqrt{A^2-B}}{2}}\pm \sqrt{\dfrac{A-\sqrt{A^2-B}}{2}}$

a)

$\sqrt{3+ \sqrt{8}}=\sqrt{\dfrac{3+\sqrt{3^2-8}}{2}}+ \sqrt{\dfrac{3-\sqrt{3^2-8}}{2}}\\=\sqrt{\dfrac{3+1}{2}}+ \sqrt{\dfrac{3-1}{2}}\\=\sqrt{2}+1$

b)

$\sqrt{11-\sqrt{21}}=\sqrt{\dfrac{11+\sqrt{11^2-21}}{2}}- \sqrt{\dfrac{11-\sqrt{11^2-21}}{2}}\\=\sqrt{\dfrac{11+10}{2}}- \sqrt{\dfrac{11-10}{2}}=\dfrac{\sqrt{21}-1}{\sqrt{2}}\\=\dfrac{\sqrt2}{2}(\sqrt{21}-1)$