Question

I will rate you brainliest

Answer 1

$\dfrac{{x+4\over2}}{{x^2-16\over3}}$

$=\dfrac{3(x+4)}{2(x+4)(x-4)}=\frac{3}{2(x-4)}$

but in original fraction, denominator can't be zero so we have to exclude x=±4

do that answer is D

Answer 2

Answer:

D) 3/2(X-4)

Step-by-step explanation:

Invert and multiply to get:

3(x+4)/2(x²-16)

factor the bottom

3(x+4)/2(x+4)(x-4)

The (x+4)’s cancel out, and you’re left with

3/2(X-4)