Question

Which of the following expressions are equivalent to 2/x^4-y-^4 choose all that apply

Answer 1

Answer:

The answer are options

$B: \frac{2}{x^2-y^2}*\frac{1}{x^2+y^2}$ and  

$D: \frac{2}{(x^2)^2-(y^2)^2}$

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:

$(x^2-y^2).(x^2-y^2) = (x^2-y^2)^2$

As we know for the expanding of the square binomials:

$(x^2-y^2)^2 = (x^2)^2 -2*x^2.y^2 + (y^2)^2 = x^4 +2*x^2y^2 + y^4$

Which is different from the denominator compared:

$x^4 -2*x^2.y^2+y^{4} \neq x^4 - y^4$

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:

$x^4 - y^4 = (x^2 - y^2)(x^2 + y^2)$

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.

$(x^{2})^{2}-(y^{2})^{2}=x^4 - y^4$

Which results in the same as the denominator compared, this let option D to be a possible answer.

Answer 2

Answer:

B and D.

Step-by-step explanation:

B.  The numerator = 2*1 = 2 and x^4 - y^4 = (x^2 - y^2)(x^2 + y^2).

D .  (x^2)^2  - (y^2)^2 = x^4 - y^4.