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(FACULDADES OBJETIVO) Qual o valor de $\phantom{X}\dfrac{\;\sqrt{9\;}\,-\,\sqrt[\Large 3]{-8\;}\,+\,(\dfrac{1}{2})^{\large 0}\;}{(-2)^{2}\,+\,\sqrt[\Large 3]{-27\;}}\phantom{X}$?

 



resposta: 6
×
(FACULDADES OBJETIVO) Qual o valor da expressão $\phantom{X}\dfrac{\;\left( 4^{{}^{\large \frac{3}{2}}}\,-\,8^{{}^{\large \frac{2}{3}}} \right)^{\large \frac{3}{2}}\;}{\;\left[2^{{}^{\large 0}}\,+\,3^{{}^{\large -1}}\centerdot 6\,-\,(\dfrac{3}{4})^{{}^{\large 0}} \right]^{\large 2}\;}\phantom{X}$?

 



resposta: 2
×
(MACKENZIE) Qual o valor de $\phantom{X}\left[ \sqrt[\LARGE 3]{\dfrac{\;(0,005)^{{}^{\Large 2}}\,\centerdot \,0,000075\;}{10}\;} \right] \div \left[ \dfrac{5\,\centerdot \,10^{{}^{\Large -4}}\,\centerdot 2^{{}^{\large -\frac{1}{3}}}}{3^{{}^{\large -\frac{1}{3}}}} \right]\phantom{X}$?

 



resposta: 1
×
Racionalizar o denominador da fração $\phantom{X}\dfrac{1}{\sqrt[\Large 3]{5}\,+\,\sqrt[\Large 3]{2}}\phantom{X}$

 



resposta:

SOMA E DIFERENÇA DE CUBOS

$\,\boxed{\;a^3\,+\,b^3\,=\,(a\,+\,b)\,\centerdot\,(a^2\,-\,ab\,+\,b^2)\,}$
$\,\boxed{\;a^3\,-\,b^3\,=\,(a\,-\,b)\,\centerdot\,(a^2\,+\,ab\,+\,b^2)\,}$

Resolução:
Devemos multiplicar o numerador e o denominador da fração pelo fator que complete a expressão do produto notável correspondente.
$\,\dfrac{1}{\sqrt[\Large 3]{5}\,+\,\sqrt[\Large 3]{2}}\,=$ $\,\dfrac{1}{\sqrt[\Large 3]{5}\,+\,\sqrt[\Large 3]{2}}\,\centerdot\,\dfrac{(\sqrt[\Large 3]{5})^2\,-\,\sqrt[\Large 3]{5}\centerdot\sqrt[\Large 3]{2}\,+\,(\sqrt[\Large 3]{2})^2}{(\sqrt[\Large 3]{5})^2\,-\,\sqrt[\Large 3]{5}\centerdot\sqrt[\Large 3]{2}\,+\,(\sqrt[\Large 3]{2})^2}\,=$ $\,\dfrac{\sqrt[\Large 3]{5^2}\,-\,\sqrt[\Large 3]{5\,\centerdot \,2\;}\,+\,\sqrt[\Large 3]{2^2}}{(\sqrt[\Large 3]{5\;})^3\,+\,(\sqrt[\Large 3]{2\;})^3}\,=$
$\,\dfrac{\;\sqrt[\Large 3]{25\;}\,-\;\sqrt[\Large 3]{10\;}\,+\,\sqrt[\Large 3]{4\;}\,}{7}\;$

×
Racionalizar o denominador da fração $\phantom{X}\dfrac{4}{\;2\,+\,\sqrt{3\,}\,+\,\sqrt{7\,}}\phantom{X}$

 



resposta:

DIFERENÇA DE QUADRADOS
$\,\boxed{\;a^2\,-\,b^2\,=\,(a\,+\,b)\,\centerdot\,(a\,-\,b)\,}$


Resolução:
Multiplicamos o numerador e o denominador da fração por $\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,$
$\,\dfrac{4}{\;2\,+\,\sqrt{3\,}\,+\,\sqrt{7\,}}\;=$ $\,\dfrac{4}{\;2\,+\,\sqrt{3\,}\,+\,\sqrt{7\,}}\,\centerdot\,\dfrac{\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,}{\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,}\,=$ $\,\dfrac{4(\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,)}{\;\left[\,2\,+\,\sqrt{3}\,+\,\sqrt{7}\,\right]\,\centerdot\,\left[\,\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,\right] \;}\,=$ $\,\dfrac{4(\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,)}{\;(2\,+\sqrt{3\,})^2\,-\,(\sqrt{7\,})^2\;}\,=$ $\dfrac{4(\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,)}{\;4\,+\,2\,\centerdot\,2\,\centerdot\,\sqrt{3\,}\,+\,3\,-\,7\;}\,=$ $\dfrac{4(\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,)}{\;4\sqrt{3\,}\;}\,=$ $\dfrac{\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,}{\;\sqrt{3\,}\;}\,=$ $\dfrac{\,2\,+\,\sqrt{3}\,-\,\sqrt{7}\,}{\;\sqrt{3\,}\;}\,\centerdot\,\dfrac{\;\sqrt{3\,}\;}{\;\sqrt{3\,}\;}\,=$ $\dfrac{\,2\sqrt{3\,}\,+\,(\sqrt{3\,})^2\,-\,\sqrt{3\,} \centerdot \sqrt{7\,}\,}{(\sqrt{3\,})^2}\,=$ $\,\dfrac{\,2\sqrt{3\,}\,+\,3\,-\,\sqrt{21\,}\,}{3}\;$
$\boxed{\,\dfrac{\,2\sqrt{3\,}\,+\,3\,-\,\sqrt{21\,}\,}{3}\;}$
×
Sendo $\;a\;$ e $\;b\;$ números reais estritamente positivos e distintos, mostrar que $\phantom{X}\dfrac{a\,-\,b}{\;\sqrt{a\,}\,-\,\sqrt{b\,}\;}\,=\,\sqrt{a\,}\,+\,\sqrt{b\,}\phantom{X}$

 



resposta:

DIFERENÇA DE QUADRADOS
$\,\boxed{\;a^2\,-\,b^2\,=\,(a\,+\,b)\,\centerdot\,(a\,-\,b)\,}$


Resolução:
$\,\dfrac{a\,-\,b}{\;\sqrt{a\,}\,-\,\sqrt{b\,}\;}\,=$ $\,\dfrac{a\,-\,b}{\;\sqrt{a\,}\,-\,\sqrt{b\,}\;}\,\centerdot \,\dfrac{\sqrt{a\,}\,+\,\sqrt{b\,}}{\;\sqrt{a\,}\,+\,\sqrt{b\,}\;}\,=$ $\,\dfrac{\;(a\,-\,b)(\sqrt{a\,}\,+\,\sqrt{b\,})\;}{(\sqrt{a\,})^2\,-\,(\sqrt{b\,})^2}\,=$ $\,\dfrac{\;(a\,-\,b)(\sqrt{a\,}\,+\,\sqrt{b\,})\;}{(a\,-\,b)}\,=\,$$\,\sqrt{a\,}\,+\,\sqrt{b\,} $

×
Os denominadores das frações abaixo são diferentes de zero. Simplifique:
a) $\,\dfrac{\;a\,+\,a^2\;}{\;2\,+\,2a\;}$

b) $\,\dfrac{\;a^3\,+\,a^2b\;}{\;a^2\,+\,2ab\,+\,b^2\;}$

 



resposta: a) a/2b) $\,\frac{a^2}{a\,+\,b}$

×
(UBERABA) Racionalizando-se o denominador da fração $\phantom{X}\dfrac{2\sqrt{3}}{\;\sqrt{5}\,-\,\sqrt{3}\;}\phantom{X}$ obtém-se:
a)
$\,\dfrac{\;\sqrt{15}\,-\,\sqrt{3}\;}{2}\,$
b)
$\,\dfrac{\;\sqrt{15}\,+\,\sqrt{3}\;}{2}\,$
c)
$2(\sqrt{15}\,+\,\sqrt{3})\,$
d)
$\,\sqrt{15}\,+\,3\,$
e)
$\,\sqrt{15}\,-\,3\,$
 
 

 



resposta: (D)
×
(MACKENZIE) $\phantom{X}\dfrac{1}{\;1\,-\,\sqrt{2}\;}\,-\,\dfrac{1}{\;\sqrt{2}\,+\,1\;}\phantom{X}$ é igual a:
a)
$\,\sqrt{2}\;\;$
b)
$\,-2\phantom{X}$
c)
$\,2\;\phantom{X}\,$
d)
$\,2(\sqrt{2}\,+\,1)\,$
e)
$\,-2\sqrt{2}\,$

 



resposta: (E)
×
(FAAP) Simplificar $\phantom{X}\dfrac{\;2\,+\,\sqrt{3}\;}{\;1\,-\,\sqrt{5}\;}\,+\,\dfrac{\;2\,-\,\sqrt{3}\;}{\;1\,+\sqrt{5}\;}\phantom{X}$

 



resposta: $\,-1\,-\,\dfrac{\sqrt{15}}{2}\,$
×
(UNB) Sejam a e b reais com |a| > |b|. O valor da expressão $\phantom{X}\dfrac{\;\sqrt{a^2\,-\,b^2\,}}{a\,-\,b}\phantom{X}$ é:
a)
1  
b)
$\,ab\,$
c)
$\,\sqrt{\,a\,+\,b\;}$ 
d)
$\,\sqrt{\;\dfrac{\;a\,+\,b\;}{a\,-\,b\;}\;}\,$
e)
nenhuma dessas

 



resposta: (E)
×
(UNB) Se $\phantom{X}P\,=\,\dfrac{1}{\,\sqrt{\,7\;}\,-\,\sqrt{\,5\;}\;}\phantom{X}$; $\phantom{X}Q\,=\,\dfrac{1}{\,\sqrt{\,8\;}\,-\,\sqrt{\,5\;}\;}\phantom{X}$; $\phantom{X}R\,=\,\dfrac{\,\sqrt{\,5\;}\,+\,\sqrt{\,8\;}\;}{3}\phantom{X}$ então:
a)
P < Q < R
b)
P > Q, Q < R
c)
P > Q > R
d)
P > Q = R
e)
P = Q < R

 



resposta: (D)
×
(FEI MAUÁ) Racionalizar o denominador da fração $\phantom{X}\dfrac{1}{\;1\,+\,\sqrt{\,2\;}\,-\,\sqrt{\,3\;}\;}\phantom{X}$

 



resposta: $\,\dfrac{\;\sqrt{\,2\;}(1\,+\,\sqrt{\,2\;}\,+\,\sqrt{\,3\;})\;}{4}\,$
×
(PUC) O número $\phantom{X}3\,+\,\sqrt{\,3\;}\,+\,\dfrac{1}{\,3\,-\,\sqrt{\,3\;}\,}\,-\,\dfrac{1}{\,3\,+\,\sqrt{\,3\;}\,}\phantom{X}$ é igual a:
a)
$\,3\,+\,\dfrac{\;3\sqrt{\,3\;}}{2}\,$
b)
$\,3\,+\,\dfrac{\;4\sqrt{\,3\;}}{3}\,$
c)
$\,3\,-\,\dfrac{\;4\sqrt{\,3\;}}{3}\,$
d)
$\,\dfrac{\;3\,+\,2\sqrt{\,3\;}}{3}\,$
e)
$\,\dfrac{\;3\,-\,2\sqrt{\,3\;}}{3/6}\,$

 



resposta: (B)
×
(UFSM) Efetuando o produto $\phantom{X}\dfrac{\;2\,-\,\sqrt{\,3\;}\;}{\sqrt{\,5\;}}\,\centerdot \,\dfrac{\;\sqrt{\,3\;}\,+\,2\;}{\sqrt{\,5\;}\,-\,1}\phantom{X}$ teremos:
a)
$\,\dfrac{1}{\;5\,+\,\sqrt{\,5\;}\;}\phantom{X}$
b)
$\,\dfrac{\;5\,+\,\sqrt{\,5\;}\;}{20}\,$
c)
$\,\dfrac{1}{\;25\,-\,\sqrt{\,5\;}\;}\,$
d)
$\,\dfrac{\;4\,-\,2\sqrt{\,3\;}}{\;2\sqrt{\,5\;}\,-\,1\;}\,$
e)
nenhuma das anteriores
 
 

 



resposta: (B)
×
(CESGRANRIO - 1982) Sendo $\phantom{X}x\;\gt\;0\phantom{X}$, com o denominador racionalizado, a razão $\phantom{X}\dfrac{\sqrt{\;x\;}}{\;\sqrt{\;x\,+\,1\;}\,+\,\sqrt{\;x\;}\;}\phantom{X}$ é igual a:
a)
$\,2x\,+\,1\,$
b)
$\,\dfrac{1}{\;x^2\,+\,2\;}\,$
c)
$\,\dfrac{x}{\;2x\,+\,1\;}\,$
d)
$\,\dfrac{\sqrt{\;x\;}}{\;2x\,+\,1\;}\,$
e)
$\,\sqrt{\;x^2\,+\,x\;}\,-\,x\,$

 



resposta: (E)
×
Racionalizar o denominador da fração $\phantom{X}\dfrac{3}{\;\sqrt[\Large 5]{8\;}\;}\phantom{X}$.

 



resposta:
Resolução:
1) notar que $\phantom{X}\sqrt[\Large 5]{8\;}\phantom{X}$ é o mesmo que $\phantom{X}\sqrt[\Large 5]{2^3\;}\phantom{X}$
2) multiplicar o numerador e o denominador da fração por $\phantom{X}\sqrt[\Large 5]{2^2\;}\phantom{X}$ , pois $\phantom{X}\sqrt[\Large 5]{2^3\;}\,\centerdot\,\sqrt[\Large 5]{2^2\;}\,=$ $\,\sqrt[\Large 5]{2^3\,\centerdot\,2^2\;}\,=\,\sqrt[\Large 5]{2^5\;}\phantom{X}$
Assim
$\;\dfrac{3}{\;\sqrt[\Large 5]{8\;}\;}\,=$ $\,\dfrac{3}{\;\sqrt[\Large 5]{2^3\;}\;}\,=$ $\,\dfrac{3}{\;\sqrt[\Large 5]{8\;}\;}\,\centerdot\,\dfrac{\;\sqrt[\Large 5]{2^2\;}\;}{\;\sqrt[\Large 5]{2^2\;}\;}\,=$ $\,\dfrac{\;3\,\centerdot\,\sqrt[\Large 5]{2^2\;}}{\;\sqrt[\Large 5]{2^5\;}\;}\,=$ $\,\dfrac{\;3\,\centerdot\,\sqrt[\Large 5]{4\;}}{\;2\;}\,$
Resposta:$\phantom{X}\dfrac{\;3\,\sqrt[\Large 5]{4\;}}{2}\phantom{X}$
×
Veja exercÍcio sobre:
radiciação
raízes
racionalização de denominadores