Question

$\sqrt{2 + \sqrt{3} } + 1 \div \sqrt{2 + \sqrt{3} }$
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Answer 1

Answer:

$\sqrt{2+\sqrt{3}}+\dfrac{1}{\sqrt{2+\sqrt{3}}}=\sqrt{2+\sqrt{3}}+\left(2-\sqrt{3}\right)\sqrt{2+\sqrt{3}}\quad \left(\mathrm{Decimal:\quad }\:2.44949\dots \right)$

Step-by-step explanation:

$\sqrt{2+\sqrt{3}}+\dfrac{1}{\sqrt{2+\sqrt{3}}}$

$\dfrac{1}{\sqrt{2+\sqrt{3}}}$

$\mathrm{Multiply\:by\:the\:conjugate}\:\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}}$

$=\dfrac{1\cdot \sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}\sqrt{2+\sqrt{3}}}$

$1\cdot \sqrt{2+\sqrt{3}}=\sqrt{2+\sqrt{3}}$

$\sqrt{2+\sqrt{3}}\sqrt{2+\sqrt{3}}$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt{a}\sqrt{a}=a$

$\sqrt{2+\sqrt{3}}\sqrt{2+\sqrt{3}}=2+\sqrt{3}$

$=2+\sqrt{3}$

$=\dfrac{\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}$

$\mathrm{Multiply\:by\:the\:conjugate}\:\dfrac{2-\sqrt{3}}{2-\sqrt{3}}$

$=\dfrac{\sqrt{2+\sqrt{3}}\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}$

$\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=1$

$=\dfrac{\left(2-\sqrt{3}\right)\sqrt{2+\sqrt{3}}}{1}$

$\mathrm{Apply\:rule}\:\dfrac{a}{1}=a$

$=\left(2-\sqrt{3}\right)\sqrt{2+\sqrt{3}}$

$=\sqrt{2+\sqrt{3}}+\left(2-\sqrt{3}\right)\sqrt{2+\sqrt{3}}$