Question

The expression (a^-1) - (b^-1)/ (a^-2)- (b^-2)
(a) b-a/ ab
(b) b+a/ ab
(c) ab/ b+a
(d) (a^2)-(b^2)/ a-b
(e) a-b

Answer 1

(a^-1) - (b^-1)/ (a^-2)- (b^-2)
\[=\frac{\frac{1}{a} - \frac{1}{b}}{\frac{1}{a^2}- \frac{1}{b^2}}\]\[=\frac{\frac{b}{ab} - \frac{a}{ab}}{\frac{b^2}{a^2b^2}- \frac{a^2}{a^2b^2}}\]\[=\frac{\frac{b-a}{ab}}{\frac{b^2-a^2}{a^2b^2}}\]\[=\frac{b-a}{ab} \times \frac{a^2b^2}{b^2-a^2}\]\[=\frac{b-a}{ab} \times \frac{a^2b^2}{(b-a)(b+a)}\]\[=...?\]

Answer 2

i got (d) is that right?

Answer 3

Not really....
Do you understand the above workings?

Answer 4

yeah. you're supposed to multiply the numerators and the denominators. i got (a^2b^3)-(a^3b^2)/ (ab^3)-(a^3b).

Answer 5

and then i simplified it down to a^2 - b^2/ a-b

Answer 6

So weird... Actually, all you need to do is to cancel the common factors there :|

Answer 7

oh and then you get (a) right?

Answer 8

wait nevermind

Answer 9

i'm confused is it (e)?

Answer 10

Not really :|

Answer 11

could u help me out?

Answer 12

Actually, which part you don't understand?

Answer 13

the cancelling of the common factors

Answer 14

Just a minute :|

Answer 15

ok

Answer 16

Assume you have the expression \[\frac{xy}{z} \times \frac{xz}{y}\]