[tex]\displaystyle\\
a=\sqrt{6-\sqrt{35}}-\sqrt{6+\sqrt{35}}\\\\
\text{Aplicam formulele:}\\\\
\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\\\\\\
\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}}[/tex]
[tex]\displaystyle\\
\text{Rezolvare:}\\
\sqrt{6-\sqrt{35}}=\sqrt{\frac{6+\sqrt{6^2-35}}{2}}-\sqrt{\frac{6-\sqrt{6^2-35}}{2}}=\\
=\sqrt{\frac{6+\sqrt{36-35}}{2}}-\sqrt{\frac{6-\sqrt{36-35}}{2}}=\\
=\sqrt{\frac{6+1}{2}}-\sqrt{\frac{6-1}{2}}=\boxed{\sqrt{\frac{7}{2}}-\sqrt{\frac{5}{2}}}\\\\
\text{Prin analogie rezulta:}\\
\sqrt{6+\sqrt{35}}=\boxed{\sqrt{\frac{7}{2}}+\sqrt{\frac{5}{2}}}[/tex]
[tex]\displaystyle\\
a=\sqrt{6-\sqrt{35}}-\sqrt{6+\sqrt{35}}=\left(\sqrt{\frac{7}{2}}-\sqrt{\frac{5}{2}}\right)-\left(\sqrt{\frac{7}{2}}+\sqrt{\frac{5}{2}}\right)=\\\\
=\sqrt{\frac{7}{2}}-\sqrt{\frac{5}{2}}-\sqrt{\frac{7}{2}}-\sqrt{\frac{5}{2}}=\\\\
=-\sqrt{\frac{5}{2}}-\sqrt{\frac{5}{2}}=-2\sqrt{\frac{5}{2}}=-\sqrt{\frac{4\times 5}{2}}=-\sqrt{2\times5}=-\sqrt{10}\\\\
\boxed{\bf a=-\sqrt{10}}\\\\
\Big(a+\sqrt{10}\Big)^{2007} = \Big(-\sqrt{10}+\sqrt{10}\Big)^{2007} =0^{2007} =\boxed{\bf 0}[/tex]
