Question

resolva:
sqrt(2) × sqrt( 2 + sqrt(2)) × sqrt( 2 +sqrt( 2+ sqrt(2))) × sqrt(2 - sqrt( 2+ sqrt(2)))


Só pra confirma aki deu = 2sqrt(3) + 4



obs: não sei coloca uma raiz dentro da outra por isso ta sqrt se alguém quiser me ensinar agradeço

Answer 1

Olá Gustavo, boa noite!

Veja como fiz:

$\\ \displaystyle \mathsf{\sqrt{2} \cdot \sqrt{2 + \sqrt{2}} \cdot \sqrt{2 + \sqrt{2 + \sqrt{2}}} \cdot \sqrt{2 - \sqrt{2 + \sqrt{2}}} =}} \\\\ \mathsf{\sqrt{2} \cdot \sqrt{2 + \sqrt{2}} \cdot \sqrt{\left ( 2 + \sqrt{2 + \sqrt{2}} \right ) \cdot \left ( 2 - \sqrt{2 + \sqrt{2}} \right )} =}} \\\\ \mathsf{\sqrt{2} \cdot \sqrt{2 + \sqrt{2}} \cdot \sqrt{4 - (2 + \sqrt{2})} =}} \\\\ \mathsf{\sqrt{2} \cdot \sqrt{2 + \sqrt{2}} \cdot \sqrt{4 - 2 - \sqrt{2}} =}} \\\\ \mathsf{\sqrt{2} \cdot \sqrt{2 + \sqrt{2}} \cdot \sqrt{2 - \sqrt{2}} =}} \\\\ \mathsf{\sqrt{2} \cdot \sqrt{(2 + \sqrt{2}) \cdot (2 - \sqrt{2})} =}} \\\\ \mathsf{\sqrt{2} \cdot \sqrt{4 - 2} =}} \\\\ \mathsf{\sqrt{2} \cdot \sqrt{2} =} \\\\ \mathsf{\sqrt{4} =} \\\\ \boxed{\mathsf{2}}$