Question

32) Letra D PASSO A PASSO0

Answer 1

Resposta:

32. d) 2

Explicação passo-a-passo:

Lembretes:

- $(a + b)(a - b) = a^{2} - b^{2}$

- $\sqrt{a}.\sqrt{b} = \sqrt{a.b}$

32.

d)

$\sqrt{2} . \sqrt{2 + \sqrt{2}} . \sqrt{2 + \sqrt{2 + \sqrt{2}}} . \sqrt{2 - \sqrt{2 + \sqrt{2}}} = \\\\\sqrt{2.(2+\sqrt{2})} . \sqrt{(2 + \sqrt{2 + \sqrt{2}}).(2 - \sqrt{2 + \sqrt{2}})} = \\\\\sqrt{4 + 2\sqrt{2}} . \sqrt{2^{2} - (\sqrt{2 + \sqrt{2}})^{2}} = \\\\\sqrt{4 + 2\sqrt{2}} . \sqrt{4 - (2 + \sqrt{2})} = \\\\\sqrt{4 + 2\sqrt{2}} . \sqrt{4 - 2 - \sqrt{2}} = \\\\\sqrt{4 + 2\sqrt{2}} . \sqrt{2 - \sqrt{2}} = \\\\\sqrt{(4 + 2\sqrt{2}).(2 - \sqrt{2})} =$

$\sqrt{4.2 - 4\sqrt{2} + 2.2\sqrt{2} - 2\sqrt{2}.\sqrt{2}} = \\\\\sqrt{8 - 4\sqrt{2} + 4\sqrt{2} - 2\sqrt{2.2}} = \\\\\sqrt{8 - 4\sqrt{2} + 4\sqrt{2} - 2\sqrt{4}} = \\\\\sqrt{8 - 2\sqrt{4}} = \\\\\sqrt{8 - 2.2} = \\\\\sqrt{8 - 4} = \\\\\sqrt{4} = \\\\2$