## anonymous 5 years ago simplify: 5 over square root of k^7

1. anonymous

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

2. anonymous

how do i simplify this?

3. anonymous

in expo notation/.

4. amistre64

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

5. anonymous

so for any problem that looks like this i take the outter number and make it the denominator of the inner #?

6. anonymous

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

7. amistre64

exactly

8. anonymous

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

9. anonymous

i can only use positive expo

10. amistre64

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 :)

11. anonymous

$2x+3 \over x$ 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,

12. amistre64

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

13. anonymous

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.

15. anonymous

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

16. anonymous

i'm terribly confused.

17. anonymous

where did the x and x-2 go?

18. amistre64

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.

19. amistre64

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

20. amistre64

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

21. amistre64

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

22. amistre64

(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