Verified answer

Resposta:

[tex]x=\sqrt{2+ \sqrt{2+\sqrt{2+\sqrt{2+....} } } }[/tex]

[tex]x=\sqrt{2+x} } }[/tex]

x²=2+x

x²-x-2=0

x'=(1+√(1+8))/2=(1+3)/2=2

x''=(1-√(1+8))/2=(1-3)/2=-1 , como não pode ser negativo

x=2

[tex]\displaystyle \sf 2=\sqrt{4}\\\\ 2=\sqrt{2+2} \\\\ 2 = \sqrt{2+\sqrt{4}}\\\\ 2 = \sqrt{2+\sqrt{2+2}} \\\\ 2 = \sqrt{2+\sqrt{2+\sqrt{4}}} \\\\ 2 = \sqrt{2+\sqrt{2+\sqrt{2+2}}} \\\\ 2 = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{4}}}} \\\\ 2=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+2}}}} \\\\ 2 = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}.}}}} \\\\\[/tex]

Podemos repetir o processo infinitas vezes.

portanto :

[tex]\sf x = \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}}} \\\\\\ \Large\boxed{\sf \ x = 2\ }\checkmark[/tex]