Answer:
The answer are options
[tex] B: \frac{2}{x^2-y^2}*\frac{1}{x^2+y^2} [/tex] and
[tex] D: \frac{2}{(x^2)^2-(y^2)^2}[/tex]
Step-by-step explanation:
As all the options are multiplication of fractions the option A cannot be an answer because the numerator multiplication is 1 and different to 2. In the case of option C, observe that if we multiply the denominators we have:
[tex] (x^2-y^2).(x^2-y^2) = (x^2-y^2)^2 [/tex]
As we know for the expanding of the square binomials:
[tex] (x^2-y^2)^2 = (x^2)^2 -2*x^2.y^2 + (y^2)^2 = x^4 +2*x^2y^2 + y^4 [/tex]
Which is different from the denominator compared:
[tex] x^4 -2*x^2.y^2+y^{4} \neq x^4 - y^4 [/tex]
Thus option B cannot be an answer either.
Noting that the denominator compared is a square of two difference by definition, therefore, can be written as:
[tex] x^4 - y^4 = (x^2 - y^2)(x^2 + y^2) [/tex]
This results in the same denominator as option B. So, option B is a possible answer.
Finally, in the denominator of option D, we can solve the exponents of this factor.
[tex] (x^{2})^{2}-(y^{2})^{2}=x^4 - y^4[/tex]
Which results in the same as the denominator compared, this let option D to be a possible answer.
