Answer :

mariamikayla
[tex]\frac{5}{\sqrt{5-2\sqrt{6}}}^{(\sqrt{5+2\sqrt{6}}}= \frac{5\sqrt{5+2\sqrt{6}}}{\sqrt{(5-2\sqrt{6})(5+2\sqrt{6})}}= \\\\\\ = \frac{5\sqrt{5+2\sqrt{5}}}{\sqrt{25-24}}= \\\\ = \frac{5\sqrt{5+2\sqrt{6}}}{1}= \\\\ =\boxed{5\sqrt{5+2\sqrt{6}}}[/tex]


[tex]5\sqrt{5+\sqrt{24}}= \\\\ A=5 \\ B=24 \\ C^2=A^2-B= 25-24=1 \ \ \ \Longrightarrow \boxed{C=1} \\\\\\ 5\sqrt{\frac{A+C}{2}+\frac{A-C}{2}}= 5\sqrt{\frac{5+1}{2}+\frac{5-1}{2}}= \\\\=5 \sqrt{\frac{6}{2}+\frac{4}{2}}= \\\\ =\boxed{\boxed{5(\sqrt{3}+\sqrt{2})}}[/tex]

Hippalectryon
[tex]5-2\sqrt{6}}=\sqrt{3}^2-2\sqrt{3}\sqrt{2}+\sqrt{2}^2=(\sqrt{3}-\sqrt{2})^2[/tex]

Hence [tex]\dfrac{1}{\sqrt{5-2\sqrt{6}}}}=\dfrac{1}{\sqrt{3}-\sqrt{2}}=\dfrac{\sqrt3+\sqrt2}{\sqrt3^2-\sqrt2^2}=\sqrt3+\sqrt2[/tex]

Therefore [tex]\boxed{\dfrac{5}{\sqrt{5-2\sqrt6}}=5(\sqrt3+\sqrt2)}[/tex]

Other Questions