Question

Racionalize as expressões a seguir.
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Answer 1

Explicação passo-a-passo:

Racionalização do denominador :

Dada as expressões :

$\mathsf{A)~\dfrac{\sqrt{5}}{6-\sqrt{2}} }$

$\mathsf{ A~=~ \dfrac{ \sqrt{5} }{6 - \sqrt{2}} \times \dfrac{ 6 + \sqrt{2} }{ 6 + \sqrt{2} } }$

$\mathsf{ A~=~\dfrac{ \sqrt{5} \Big( 6 + \sqrt{2} \Big) }{ \Big( 6 - \sqrt{2} \Big) \Big( 6 + \sqrt{2} \Big) } }$

Note que :

(a - b)(a + b) = a² - b² , então vamos ter :

$\mathsf{A~=~ \dfrac{ 6\sqrt{5} + \sqrt{5.2} }{ 6^2 - (\sqrt{2})^2} }$

$\mathsf{A~=~ \dfrac{ 6\sqrt{5} + \sqrt{10} }{36 - 2}~=~\red{ \dfrac{ 6\sqrt{5} + \sqrt{10} }{34} } }$

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$\mathsf{B)~~ \dfrac{2}{5 - \sqrt{3}} }$

$\mathsf{ B~=~\dfrac{2}{ 5 - \sqrt{3} } \times \dfrac{ 5 + \sqrt{3}}{ 5 + \sqrt{3}} }$

$\mathsf{ B~=~ \dfrac{ 2\Big(5 + \sqrt{3} \Big)}{\Big( 5 - \sqrt{3} \Big) \Big( 5 + \sqrt{3} \Big) } }$

$\mathsf{B~=~ \dfrac{10 + 2\sqrt{3}}{5^2 - (\sqrt{3})^2} ~=~\dfrac{ 10 + 2\sqrt{3} }{25 - 3 } }$

$\mathsf{\blue{B~=~ \dfrac{10 + 2\sqrt{3} }{22} } }$

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$\mathsf{ C)~~ \dfrac{4\sqrt{3}}{ \sqrt{5} + \sqrt{2}} }$

$\mathsf{ C~=~ \dfrac{4\sqrt{3}}{ \sqrt{5} + \sqrt{2} } \times \dfrac{ \sqrt{5}- \sqrt{2} }{ \sqrt{5} - \sqrt{2} } }$

$\mathsf{C~=~ \dfrac{ 4\sqrt{3 . 5 } + 4\sqrt{3 . 2} }{ (\sqrt{5} )^2 - (\sqrt{2})^2 }~=~ \dfrac{ 4( \sqrt{8} + \sqrt{6} ) }{ 5 - 2 } }$

$\mathsf{ \green{ B~=~ \dfrac{4( 2\sqrt{2} + \sqrt{6} )}{3} } }$

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$\mathsf{ D)~~ \dfrac{ 7}{ \sqrt{3} - \sqrt{2} } }$

$\mathsf{ D~=~ \dfrac{7}{ \sqrt{3} - \sqrt{2} } \times \dfrac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } }$

$\mathsf{ D~=~ \dfrac{7( \sqrt{3} + \sqrt{2} )}{ (\sqrt{3} - \sqrt{2} )( \sqrt{3} + \sqrt{2} ) } }$

$\mathsf{ D~=~ \dfrac{7( \sqrt{3} + \sqrt{2} )}{(\sqrt{3})^2 - (\sqrt{2})^2}~=~\dfrac{ 7( \sqrt{3} + \sqrt{2} )}{ 3 - 2 } }$

$\mathsf{ \red{ D~=~ 7( \sqrt{3} + \sqrt{2} ) } }$ ✅✅

Espero ter ajudado bastante!)

Answer 2

Oie, Td Bom?!

a)

$\frac{ \sqrt{5} }{6 - \sqrt{2} }$

$\frac{ \sqrt{5} }{6 - \sqrt{2} } \times \frac{6 + \sqrt{2} }{6 + \sqrt{2} }$

$\frac{ \sqrt{5} \times (6 + \sqrt{2} )}{(6 - \sqrt{2}) \times (6 + \sqrt{2} ) }$

$\frac{6 \sqrt{5} + \sqrt{10} }{36 - 2}$

$\frac{6 \sqrt{5} + \sqrt{10} }{34}$

b)

$\frac{2}{5 - \sqrt{3} }$

$\frac{2}{5 - \sqrt{3} } \times \frac{5 + \sqrt{3} }{5 + \sqrt{3} }$

$\frac{2(5 + \sqrt{3} )}{(5 - \sqrt{3}) \times (5 + \sqrt{3} ) }$

$\frac{2(5 + \sqrt{3} )}{25 - 3}$

$\frac{2(5 + \sqrt{3}) }{22}$

$\frac{5 + \sqrt{3} }{11}$

c)

$\frac{4 \sqrt{3} }{ \sqrt{5} + \sqrt{2} }$

$\frac{4 \sqrt{3} }{ \sqrt{5} + \sqrt{2} } \times \frac{ \sqrt{5} - \sqrt{2} }{ \sqrt{5} - \sqrt{2} }$

$\frac{4 \sqrt{3} \times ( \sqrt{5} - \sqrt{2} )}{( \sqrt{5} + \sqrt{2}) \times ( \sqrt{5} - \sqrt{2} )}$

$\frac{4 \sqrt{15} - 4 \sqrt{6} }{5 - 2}$

$\frac{4 \sqrt{15} - 4 \sqrt{6} }{3}$

d)

$\frac{7}{ \sqrt{3} - \sqrt{2} }$

$\frac{7}{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

$\frac{7( \sqrt{3} + \sqrt{2}) }{( \sqrt{3} - \sqrt{2} ) \times ( \sqrt{3} + \sqrt{2} )}$

$\frac{7( \sqrt{3} + \sqrt{2} )}{3 - 2}$

$\frac{7( \sqrt{3} + \sqrt{2}) }{1}$

$7( \sqrt{3} + \sqrt{2} )$

$7 \sqrt{3} + 7 \sqrt{2}$

Att. Makaveli1996