+0

# sqrt3(x^2/yz^5)z

0
739
4
+33

sqrt3(x^2/yz^5)z

Jan 7, 2015

#3
+118623
+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
+23247
+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
+33
0

what happened with the squart3

Jan 7, 2015
#3
+118623
+10

Is this your intended question Alisson?

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

Melody Jan 7, 2015
#4
+118623
+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