Question

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Answer 1

Resposta:

Racionalizando

a)

$\dfrac{3}{\sqrt{2} }=\dfrac{3.\sqrt{2} }{\sqrt{2} .\sqrt{2} }=\dfrac{3\sqrt{2} }{\sqrt{4} }=\dfrac{3\sqrt{2} }{2}$

b)

$\dfrac{1}{\sqrt{5} }=\dfrac{\sqrt{5} }{\sqrt{5} .\sqrt{5} }=\dfrac{\sqrt{5} }{\sqrt{25} }=\dfrac{\sqrt{5} }{5}$

c)

$\dfrac{5}{\sqrt{14} }=\dfrac{5.\sqrt{14} }{\sqrt{14} .\sqrt{14} }=\dfrac{5\sqrt{14} }{\sqrt{14^2} }=\dfrac{5\sqrt{14} }{14}$

d)

$\dfrac{12}{\sqrt{6} }=\dfrac{12\sqrt{6} }{\sqrt{6} .\sqrt{6} }=\dfrac{12\sqrt{6} }{\sqrt{6^2} }=\dfrac{12\sqrt{6} }{6}=2\sqrt{6}$

e)

$\dfrac{\sqrt{12} }{\sqrt{7} }=\dfrac{2\sqrt{3} }{\sqrt{7} }=\dfrac{2\sqrt{3} .\sqrt{7} }{\sqrt{7} .\sqrt{7} }=\dfrac{2\sqrt{21} }{\sqrt{7^2} }=\dfrac{2\sqrt{21} }{7}$

f)

$\dfrac{3}{\sqrt{8} }=\dfrac{3}{2\sqrt{2} }=\dfrac{3\sqrt{2} }{2.\sqrt{2} .\sqrt{2} }=\dfrac{3\sqrt{2} }{2.\sqrt{4} }=\dfrac{3\sqrt{2} }{2.2}=\dfrac{3\sqrt{2} }{4}$