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$\sqrt{32}-\sqrt8+11\sqrt{72}=$
$2\sqrt{108}-2\sqrt{27}+12\sqrt{12}=$

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$\dfrac{2}{1+\sqrt3}=$
$\dfrac{\sqrt3}{\sqrt3-\sqrt2}=$

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$\sqrt{5-2\sqrt6}=$
$\left(\dfrac1{\sqrt{14}-\sqrt{13}}\right)^2+\dfrac1{\sqrt{14}-\sqrt{13}}\cdot\dfrac1{\sqrt{14}+\sqrt{13}}+\left(\dfrac1{\sqrt{14}+\sqrt{13}}\right)^2=$
$\dfrac{\sqrt{3+\sqrt5}+\sqrt{3-\sqrt5}}{\sqrt{3-\sqrt8}}-\dfrac{4}{\sqrt{8-2\sqrt{15}}}=$

Řešení	Vybrat Ukázat
1. $\sqrt{5-2\sqrt6}=\sqrt{3-2\sqrt3\cdot\sqrt2+2}=\sqrt{(\sqrt3-\sqrt2)^2}=\sqrt3-\sqrt2$ 3. a) $\sqrt{3+\sqrt5}+\sqrt{3-\sqrt5}=\sqrt{\frac{6+2\sqrt5}2}+\sqrt{\frac{6-2\sqrt5}2}=\sqrt{\frac{5+2\sqrt5\cdot1+1}{2}}+\sqrt{\frac{5-2\sqrt5\cdot1+1}{2}}=$ $\sqrt{\frac{(\sqrt5+1)^2}{2}}+\sqrt{\frac{(\sqrt5-1)^2}{2}}=\frac{\sqrt5+1}{\sqrt2}+\frac{\sqrt5-1}{\sqrt2}=\sqrt{10}$ b) $\sqrt{3-\sqrt8}=\sqrt{2-2\sqrt2\cdot1+1}=\sqrt{(\sqrt2-1)^2}=\sqrt2-1$ c) $\sqrt{8-2\sqrt{15}}=\sqrt{5-2\sqrt5\sqrt3+3}=\sqrt{(\sqrt5-\sqrt3)^2}=\sqrt5-\sqrt3$ $\dfrac{\sqrt{3+\sqrt5}+\sqrt{3-\sqrt5}}{\sqrt{3-\sqrt8}}-\dfrac{4}{\sqrt{8-2\sqrt{15}}}=\dfrac{\sqrt{10}}{\sqrt2-1}-\dfrac{4}{\sqrt5-\sqrt3}=$ $\sqrt{20}+\sqrt{10}-\dfrac{4(\sqrt5+\sqrt3)}{2}=\sqrt{10}-2\sqrt{3}$

Řešení

Ukázat

1. $\sqrt{5-2\sqrt6}=\sqrt{3-2\sqrt3\cdot\sqrt2+2}=\sqrt{(\sqrt3-\sqrt2)^2}=\sqrt3-\sqrt2$

a) $\sqrt{3+\sqrt5}+\sqrt{3-\sqrt5}=\sqrt{\frac{6+2\sqrt5}2}+\sqrt{\frac{6-2\sqrt5}2}=\sqrt{\frac{5+2\sqrt5\cdot1+1}{2}}+\sqrt{\frac{5-2\sqrt5\cdot1+1}{2}}=$

$\sqrt{\frac{(\sqrt5+1)^2}{2}}+\sqrt{\frac{(\sqrt5-1)^2}{2}}=\frac{\sqrt5+1}{\sqrt2}+\frac{\sqrt5-1}{\sqrt2}=\sqrt{10}$

b) $\sqrt{3-\sqrt8}=\sqrt{2-2\sqrt2\cdot1+1}=\sqrt{(\sqrt2-1)^2}=\sqrt2-1$

c) $\sqrt{8-2\sqrt{15}}=\sqrt{5-2\sqrt5\sqrt3+3}=\sqrt{(\sqrt5-\sqrt3)^2}=\sqrt5-\sqrt3$

$\dfrac{\sqrt{3+\sqrt5}+\sqrt{3-\sqrt5}}{\sqrt{3-\sqrt8}}-\dfrac{4}{\sqrt{8-2\sqrt{15}}}=\dfrac{\sqrt{10}}{\sqrt2-1}-\dfrac{4}{\sqrt5-\sqrt3}=$

$\sqrt{20}+\sqrt{10}-\dfrac{4(\sqrt5+\sqrt3)}{2}=\sqrt{10}-2\sqrt{3}$

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