+0

0
334
7
+2448

Nov 7, 2018

#1
+342
+1

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

.
Nov 7, 2018
edited by Dimitristhym  Nov 7, 2018
edited by Dimitristhym  Nov 7, 2018
edited by Dimitristhym  Nov 7, 2018
#2
+2448
+1

Is that completely simplified?

RainbowPanda  Nov 7, 2018
#3
+342
+2

Yes you can do and this :

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

Dimitristhym  Nov 7, 2018
#4
+111329
+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 ]  ÷  [ ( 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

Nov 7, 2018
#5
+2448
+1

Thank you! ^-^

RainbowPanda  Nov 7, 2018
#6
+342
+1

Nice CPhill!

Dimitristhym  Nov 7, 2018
#7
+111329
0

Thanks, Dimitristhym.......!!!!

CPhill  Nov 7, 2018