Question

racionalize as fracões a seguir
​

Answer 1

Resposta:

a)

$\dfrac{ \sqrt{4} }{ \sqrt{5} } = \dfrac{2}{ \sqrt{5} } \times \dfrac{ \sqrt{5} }{ \sqrt{5} } = \\ \\ \dfrac{2 \sqrt{5} }{ \sqrt{5 \times 5} } = \frac{2 \sqrt{5} }{ \sqrt{25} } = \\ \\ \blue{ \dfrac{2 \sqrt{5} }{5} }$

b)

$\dfrac{ \sqrt{3} }{ \sqrt{2} } \times \dfrac{ \sqrt{2} }{ \sqrt{2} } = \\ \\ \frac{ \sqrt{3 \times 2} }{ \sqrt{2 \times 2} } = \frac{ \sqrt{6} }{ \sqrt{4} } = \\ \\ \blue{ \dfrac{ \sqrt{6} }{2} }$

c)

$\dfrac{ \sqrt{2} }{ \sqrt{3} } \times \dfrac{ \sqrt{3} }{ \sqrt{3} } = \\ \\ \frac{ \sqrt{2 \times 3} }{ \sqrt{3 \times 3} } = \frac{ \sqrt{6} }{ \sqrt{9} } = \\ \\ \blue{\dfrac{ \sqrt{6} }{3} }$

d)

$\dfrac{5 \sqrt{7} }{ \sqrt{2} + 1 } \times \dfrac{( \sqrt{2} - 1)}{( \sqrt{2} - 1)} = \\ \\ \dfrac{5 \sqrt{14} - 5 \sqrt{7} }{ {( \sqrt{2} })^{2} - {(1)}^{2} } = \dfrac{5 \sqrt{14} - 5 \sqrt{7} }{2 - 1} = \\ \\ \dfrac{5 \sqrt{14} - 5 \sqrt{7} }{1} \\ \\ \blue{5 \sqrt{14} - 5 \sqrt{7} }$

e)

$\dfrac{4 \sqrt{2} }{ \sqrt{3} } \times \dfrac{ \sqrt{3} }{ \sqrt{3} } = \\ \\ \dfrac{4 \sqrt{2 \times 3} }{ \sqrt{3 \times 3} } = \dfrac{4 \sqrt{6} }{ \sqrt{9} } = \\ \\ \blue{ \dfrac{4 \sqrt{6} }{3} }$

f)

$\dfrac{6}{ \sqrt{3} } \times \dfrac{ \sqrt{3} }{ \sqrt{3} } = \dfrac{6 \sqrt{3} }{ \sqrt{3 \times 3} } = \\ \\ \dfrac{6 \sqrt{3} }{ \sqrt{9} } = \dfrac{6 \sqrt{3} }{3} = \\ \\ \blue{2 \sqrt{3} }$

g)

$\dfrac{5 \sqrt{2} }{2 \sqrt{5} } \times \dfrac{2 \sqrt{5} }{2 \sqrt{5} } = \\ \\ \dfrac{5 \times 2 \sqrt{2 \times 5} }{2 \times 2 \sqrt{5 \times 5} } = \dfrac{10 \sqrt{10} }{4 \sqrt{25} } = \\ \\ \dfrac{10 \sqrt{10} }{4(5)} = \dfrac{10 \sqrt{10} \div 10}{20 \div 10} = \\ \\ \blue{\dfrac{ \sqrt{10} }{2} }$

h)

$\dfrac{ \sqrt{2} + \sqrt{5} }{ \sqrt{5} } \times \dfrac{ \sqrt{5} }{ \sqrt{5} } = \\ \\ \frac{ \sqrt{2 \times 5} + \: \sqrt{5 \times 5} }{ \sqrt{5 \times 5} } = \\ \\ \frac{\sqrt{10} + \sqrt{25} }{ \sqrt{25} } = \\ \\ \blue{ \dfrac{ \sqrt{10} + 5}{5} }$

Bons Estudos!