3) Fatore o radicando e simplifique os radicais: Resolvido: sqrt (32)=sqrt (2^5)=sqrt (2^4cdot 2)= 2^2sqrt (2)=4sqrt (2) a) sqrt (8) d) sqrt [4](32) b) sqrt (27) e) sqrt (50) c) sqrt [3](81) f) sqrt (80) 1) Calcule:

a) $$\sqrt{8} = \sqrt{2^3} = \sqrt{2^2 \cdot 2} = 2\sqrt{2}$$
b) $$\sqrt{27} = \sqrt{3^3} = \sqrt{3^2 \cdot 3} = 3\sqrt{3}$$
c) $$\sqrt[3]{81} = \sqrt[3]{3^4} = \sqrt[3]{3^3 \cdot 3} = 3\sqrt[3]{3}$$
d) $$\sqrt[4]{32} = \sqrt[4]{2^5} = \sqrt[4]{2^4 \cdot 2} = 2\sqrt[4]{2}$$
e) $$\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$$
f) $$\sqrt{80} = \sqrt{16 \cdot 5} = 4\sqrt{5}$$

1) Calcule: $$\sqrt{32} = 4\sqrt{2}$$