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25*x^0 ((y^-1 )/ (5x^-1)) * 5^-1y^2

 Nov 20, 2014

Best Answer 

 #1
avatar+23254 
+5

x^0  =  1

Y^-1  =  1/y

5x^-1  =  5/x

5^-1  =  1/5

25*x^0 ((y^-1 )/ (5x^-1)) * 5^-1y^2

= (25)*(1) ( (1/y) / (5/x) )*(1/5)y^2

    (1/y) / (5/x)  =  (1/y) * (x/5)  =  x/(5y)

= (25)*(1) ( x/(5y) )*(1/5)y^2

= 25 * ( x/(5y) )*(1/5)y^2

= [25xy^2] / [25y]

= xy

 Nov 20, 2014
 #1
avatar+23254 
+5
Best Answer

x^0  =  1

Y^-1  =  1/y

5x^-1  =  5/x

5^-1  =  1/5

25*x^0 ((y^-1 )/ (5x^-1)) * 5^-1y^2

= (25)*(1) ( (1/y) / (5/x) )*(1/5)y^2

    (1/y) / (5/x)  =  (1/y) * (x/5)  =  x/(5y)

= (25)*(1) ( x/(5y) )*(1/5)y^2

= 25 * ( x/(5y) )*(1/5)y^2

= [25xy^2] / [25y]

= xy

geno3141 Nov 20, 2014

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