+0  
 
0
43
7
avatar+2072 

please help asap, ty

RainbowPanda  Nov 7, 2018
 #1
avatar+316 
+2

\(\frac{\frac{a^2}{3^2}}{\frac{3^2-a^2}{3a^2}}=\frac{3a^2a^2}{3^23^2-3^2a^2}=\frac{a^4}{3^3-3a^2}\)

Dimitristhym  Nov 7, 2018
edited by Dimitristhym  Nov 7, 2018
edited by Dimitristhym  Nov 7, 2018
edited by Dimitristhym  Nov 7, 2018
 #2
avatar+2072 
+1

Is that completely simplified?

RainbowPanda  Nov 7, 2018
 #3
avatar+316 
+3

Yes you can do and this : 

\(\frac{a^4}{3(3^2-a^2)}\)

Dimitristhym  Nov 7, 2018
 #4
avatar+91360 
+3

Lets work with this, first

 

3                  1

____    -     ___

a^2             3

 

 

Cross-multiply ans keep the same sign as between the fractions

3*3 - 1*a^2   =  9 - a^2

Take the product of the denominators 

3a^2

 

So   we have

 

( 9 - a^2) 

________

    3a^2

 

So...puitting this together, we have

 

[ a^2 / 9 ]  รท  [ ( 9 - a^2) / ( 3a^2) ]

 

Invert the second fraction and multiply

 

a^2               3a^2                      3a^4

___   *         ______    =           _______   

9                  9 - a^2                  9(9 - a^2)

 

Cancel  the 3 in the numerator and the 9 in the denominator  and we are left with

 

a^4                        a^4

_______     =      ________

3(9 - a^2)           27 - 3a^2

 

 

 

 

 

cool cool cool

CPhill  Nov 7, 2018
 #5
avatar+2072 
+1

Thank you! ^-^ 

RainbowPanda  Nov 7, 2018
 #6
avatar+316 
+2

Nice CPhill! 

Dimitristhym  Nov 7, 2018
 #7
avatar+91360 
0

Thanks, Dimitristhym.......!!!!

 

 

cool cool cool

CPhill  Nov 7, 2018

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