2/(x-2)(x+2) - 1/x(x+2x)?

Answer 1

(5x^2 - 4) / [(x-2)(x+2)(3x^2)]

Answer 2

It's possible that you mean
2/[(x-2)(x+2)] - 1/[x(x^2+2x)]
= 2 / [(x-2)(x+2)] - 1 / [x^2(x+2)]
So the common denominator will be x^2(x-2)(x+2):
= 2x^2 / [x^2(x-2)(x+2)] - 1(x-2) / [x^2(x-2)(x+2)]
= (2x^2 - x + 2) / [x^2(x-2)(x+2)]
The numerator is irreducible (discriminant is b^2 - 4ac = 1 - 4.2.2 = -15 < 0) so this is the best we can do.

It's also possible that you mean
2/[(x-2)(x+2)] - 1/[x(x+2)]
So the common denominator will be x(x-2)(x+2):
= 2x / [x(x-2)(x+2)] - 1(x-2) / [x(x-2)(x+2)]
= (2x - x + 2) / [x^2(x-2)(x+2)]
= (x+2) / [x^2(x-2)(x+2)]
= 1 / [x^2(x-2)]

Since this works out nicely I expect this is the correct version of the problem.

Answer 3

in order to add or subt fractions u must get a common demoninator. the denominator of the second term is xtimes3x which = 3x^2.
2*3x^2/(x-2)(x+2)*3x^2 -1*(x-2)(x+2)/(x-2)(x+2)*3x^2: now you can combine the numerators since the denominators are the same.
6x^2 -x^2-4/(x-2)(x+2)*3x^2:simplify further:
5x^2-4/(x-2)(x+2)*3x^2: if u multiply out the denominator then:
5x^2-4/3x^4-12x^2
are u sure u copied this problem correctly?