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0
730
4
avatar+33 

sqrt3(x^2/yz^5)z

 Jan 7, 2015

Best Answer 

 #3
avatar+118604 
+10

Is this your intended question Alisson?

 

$$\\\sqrt3\times \dfrac{x^2}{yz^5}\times z\\\\\\
=\sqrt3\times \dfrac{x^2}{yz^4}\\\\\\
=\dfrac{\sqrt{3}\;x^2}{yz^4}\\\\\\$$

 Jan 7, 2015
 #1
avatar+23245 
+10

[ x² / (yz5) ](1/3) · z  

To rationalize the denominator under the radical sign, multiply the numerator and the denominator by y²z:

[ x²·y²z / (yz5·y²z) ](1/3) · z  

=  [ x²·y²z / (y3z6) ](1/3) · z 

=  (x²·y²z)1/3 / (yz²)  · z 

=  (x²·y²z)1/3 / (yz)

 Jan 7, 2015
 #2
avatar+33 
0

what happened with the squart3

 Jan 7, 2015
 #3
avatar+118604 
+10
Best Answer

Is this your intended question Alisson?

 

$$\\\sqrt3\times \dfrac{x^2}{yz^5}\times z\\\\\\
=\sqrt3\times \dfrac{x^2}{yz^4}\\\\\\
=\dfrac{\sqrt{3}\;x^2}{yz^4}\\\\\\$$

Melody Jan 7, 2015
 #4
avatar+118604 
+5

or do you mean

 

$$\sqrt[3]{x^2/yz^5)}\times z\\\\\\
=\sqrt[3]{\dfrac{x^2}{yz^5}}\times z\\\\\\
=\dfrac{x^{2/3}}{y^{1/3}z^{5/3}}\times z\\\\\\
=\dfrac{x^{2/3}}{y^{1/3}z^{3/3}z^{2/3}}\times z\\\\\\
=\dfrac{x^{2/3}}{y^{1/3}z^{2/3}}\\\\\\
=\sqrt[3]{\dfrac{x^{2}}{yz^{2}}}\\\\\\$$

 

I think Geno's answer is another presentation of this.   I am sure it is correct to.

 Jan 7, 2015

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