Question

Ajutor

Answer 1

 

$\displaystyle\bf\\1)~~Demonstrati~ca:~~~\sqrt{3+\sqrt{5}}=\frac{\sqrt{5}+1}{\sqrt{2}}\\\\\textbf{Folosim formula radicalilor compusi.}\\\\\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$

.

$\displaystyle\bf\\Rezolvare:\\\\\sqrt{3+\sqrt{5}}=\sqrt{\frac{3+\sqrt{3^2-5}}{2}}+\sqrt{\frac{3-\sqrt{3^2-5}}{2}}=\\\\=\sqrt{\frac{3+\sqrt{9-5}}{2}}+\sqrt{\frac{3-\sqrt{9-5}}{2}}=\\\\=\sqrt{\frac{3+\sqrt{4}}{2}}+\sqrt{\frac{3-\sqrt{4}}{2}}=\\\\=\sqrt{\frac{3+2}{2}}+\sqrt{\frac{3-2}{2}}=\\\\=\sqrt{\frac{5}{2}}+\sqrt{\frac{1}{2}}=\\\\={\frac{\sqrt{5}}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\boxed{\bf\frac{\sqrt{5}+1}{\sqrt{2}}}$