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Question Number 109222 by bemath last updated on 22/Aug/20

$$\frac{\dots \mathrm{♭}em\mathcal{ATH}\dots }{\cong \cong \cong \cong \cong \cong }$$$$\sqrt{5+\sqrt{10}}=x.(\sqrt{5+\sqrt{15}}+\sqrt{5-\sqrt{15}})$$$$x=?$$

Answered by bobhans last updated on 22/Aug/20

$$\frac{\mathrm{\_}\mathrm{♭}o\mathrm{♭}\mathcal{HAN}\varsigma \mathrm{\_}}{▹◃}$$$$(\sqrt{5+\sqrt{15}}+\sqrt{5-\sqrt{15}})={\left(\sqrt{\sqrt{5+\sqrt{15}}+\sqrt{5-\sqrt{15}}}\right)}^2$$$$=\sqrt{5+\sqrt{15}+5-\sqrt{15}+2\sqrt{25-15}}$$$$=\sqrt{10+2\sqrt{10}}=\sqrt{2(5+\sqrt{10})}=\sqrt{2}\sqrt{5+\sqrt{10}}$$$$sowegot$$$$\sqrt{5+\sqrt{10}}=x.\sqrt{2}.\sqrt{5+\sqrt{10}}$$$$\Rightarrow 1=x\sqrt{2}\Rightarrow x=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$$

Commented by bemath last updated on 22/Aug/20

$$yes...thankyou$$

Commented by bemath last updated on 22/Aug/20

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