find the sum

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

the sum is a fraction with __ in the numerator and in the __ in the denominator.

find the sum 2bba2 and 2ab2a2 the sum is a fraction with in the numerator and in the in the denominator class=

Respuesta :

Answer:

Numerator = 2(b^2+a^2)    or equivalently 2b^2+2a^2

Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3

Step-by-step explanation:

Let

S = 2b/(b+a)^2 + 2a/(b^2-a^2)   factor denominator

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

= 1/(b+a) ( 2b/(b+a) + 2a/(b-a))    find common denominator

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

= 1/(b+a)(2b^2-2ab+2ab+2a^2)/((b+a)(b-a))  simplify & factor

= 2/(b+a)(b^2+a^2)/((b+a)(b-a))  simplify & rearrange

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

Numerator = 2(b^2+a^2)    or equivalently 2b^2+2a^2

Denominator = (b+a)^2*(b-a), or equivalently b^3+ab^2-a^2b0-a^3