Factorise and simplify 2b-2a ÷ a²-b²?

2b-2a ÷ a²-b²=2(b-a)/[(a-b)*(a+b)]
=-2(a-b)/[(a-b)*(a+b)]
=-2/(a+b)

-2(a-b)\(a-b)(a+b)
(a-b) cancels with (a-b)
-2\(a+b) or 2\(-b-a)

2b - 2a ÷ a² - b² = -2(a - b) / [(a - b)(a + b)]

= -2 / (a + b)

Note: I factored 2b - 2a as -2(a - b) because I wanted (a - b) in the numerator to cancel with the (a - b) in the denominator.

I used the identity a² - b² = (a - b)(a + b) [that is the difference of two squares] to factor the denominator.

Assuming that you mean :-
2 (b - a)
----------------
a ²- b ²

2(b - a)
------------------
( a - b ) ( a + b )

- 2 (a - b)
----------------------
( a - b ) ( a + b )

- 2
-------------
( a + b )

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

2b - 2a ÷ a² - b²
= 2(b - a) ÷ (a + b)(a - b)
= -2(a - b) ÷ (a + b)(a - b)
= -2 ÷ (a + b)
= -2/(a + b)

= 2(b-a)/ (a-b)(a+b) = 2* (-1)/ ( a+b) =- 2/a+b

case 1:
(2b - 2a)/(a² - b²)
=2(b-a)/(a+b)(a-b)
=2(b-a)/-(a+b)(b-a)
=2/-(a+b)
= -2/a+b
case 2:
2b + (-2a/a²) - b²
= 2b - b² + (-2a/a²)
= b(2-b) - 2/a

= (2b - 2a) / (a+b) x (a-b)
= 2 x (b - a) / (a+b) x (a-b)
= 2 x (a - b) / (a+b) x (a-b)
= 2 / (a+b)

a^2-b^2= (a-b)*(a+b) and 2b-2a= 2*(b-a)

so simplifying by (b-a)

you obtain 2/(a+b)