simplify: 5 over square root of k^7

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simplify: 5 over square root of k^7

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[\sqrt[5]{k ^{7}}\]
how do i simplify this?
in expo notation/.

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Other answers:

k to the power of (7/5) is the best you can do with it. k^(7/5)
so for any problem that looks like this i take the outter number and make it the denominator of the inner #?
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
move everything to the top except for the (15) and make their exponents the opposite sign: (63/15) (m^-2)(m^-6) (v^3)(v^5) When you multiply like bases the exponents add together: (63/15) (m^(-2-6)) (v^(3+5)) (63/15) m^-8 v^8 ; now everything with a negative exponent gets dropped to the bottom and reverse the sign. (63 v^8) / (15 m^8) reduce your fraction to: (12 v^8)/ (5 m^8) as long as I havent any typos, that should be it :)
\[2x+3 \over x \] MINUS \[x-2 \over x+1\] YOU ARE A lifesaver. thanks so much if you have time here is one more that i am stuck on,
glorified fractions is all these are. which means we have to get like denominators. The like denominator with simply be (x)(x+1) (2x+3)(x+1) = 2x^2 +5x +3 (over (x)(x+1) of course) and (x-2)(x) = x^2 -2x (over (x)(x+1)) (2x^2 +5x +3) - (x^2 -2x) = 2x^2 +5x +3 -x^2 +2x lets clean that up to read: (x^2 +7x +3) over (x^2 +x) That should work
i'm lost...
back to your original problem there is a conflict between what you state and the way you expressed by notation. You say 5 over the square root of k^7l, but you indicate the fifth root (not square root) of k^7. I didn't follow the rest as I want to know what you are trying to simplify.
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?
when dealing with fraction; the only way to add or subtract them together is to get a common denominator. the common denominator of (x) and (x+1) is simply (x)(x+1) so we have to modify these fractions by muliplying them by either (x+1) over (x+1) ... OR (x)/(x) to get them to have common denominators.
(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)
(2x^2 +5x +3) - (x^2 -2x) is the same as: 2x^2 +5x +3 -x^2 +2x so combine like terms now (2-1)x^2 +(5+2)x +3 = x^2 +7x +3 x^2 +7x +3 ---------- (x)(x+1) x^2 +7x +3 ---------- x^2 + x

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