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\[\sqrt[5]{k ^{7}}\]

how do i simplify this?

in expo notation/.

k to the power of (7/5) is the best you can do with it.
k^(7/5)

ie 7 square root k^4 woudl be k^ 4/7

exactly

so here is one\[63m^{-2}v ^{3} \over 15m ^{6}v ^{-5}\] more if you don't mind....

i can only use positive expo

i'm lost...

so you had the (2x+3)(x+1) AND THEN you had that equal something, where did you get that?

i'm terribly confused.

where did the x and x-2 go?

(2x + 3) (x+1) (2x^2 +5x +3)
------- ----- = ------------
(x) (x+1) (x)(x+1)

(x-2) (x) (x^2 -2x)
----- -- = -------
(x+1) (x) (x)(x+1)

(2x^2 +5x +3) - (x^2 -2x)
------------------------
(x)(x+1)