how would I simplify this:(a/2b - 2b/a) / (1/b^2 - 1/2a)?

Question

(a/2b - 2b/a) / (1/b^2 - 1/2a)

Answer 1

Both numerator and denominator of the large fraction contain fractions. The common denominator of all four of those is 2ab^2. Multiply them all with this factor. You get

(a^2 b - 4 b^3) / (2 a - b^2)

You can factor out one b yet

b (a^2 - 4 b^2) / (2 a - b^2)

and for the real fancy work, factor:

b (a - 2 b) (a + 2b) / (2 a - b^2)

That's about as far as it goes :)

Answer 2

Start with the top: a/2b-2b/a. By finding a common denominator, you get: a^2/2ab-[4(b^2)]/2ab. By adding, you get: [a^2-4(b^2)]/2ab. That is the numerator of your fraction. Then work with the bottom: 1/b^2-2b/a. By finding a common denominator, you get: 2a/2a(b^2)-b^2/2a(b^2). By adding, you get: (2a-b^2)/[2a(b^2)]. This is the denominator of your fraction. You can further simplify by inverting the denomiator and multiplying it by the numerator.
{[a^2-4(b^2)]/2ab}TIMES{(2a-b^2)/{2a(b^2)]
2ab appears in the denominator of the 1st fraction and the numerator of the 2nd fraction, thus cancels out. Your product is: [b(a^2)-4(b^3)]/(2a-b^2). That is as simple as I could get it.

Answer 3

Mulitply the entire ratio by the ratio of the LCD:

In other words, multiply by 1 or (2ab^2)/(2ab^2)

(2ab^2)/(2ab^2) * (a/2b - 2b/a) / (1/b^2 - 1/2a)=

(a^2b - 4b^2)/(2a-b^2)

the guy above me is wrong. a/2b - 2b/a = a^2/2ab - 4b^2/2ab = (a^2-4b^2)/2ab NOT (a^2-b^2)/2ab

Answer 4

a/2b-2b/a=a^2-b^2/2ab
1/b^2-1/2a=2a-b^2/2ab^2
now simplifying
b(a^2-b^2)/(2a-b^2)

Answer 5

1.53535345353^3