Question

What is the simplified form of 4x/x^2-4 divided by 8x^2/x^2+4x+4 ?

Answer 1

\[\frac{4x}{x^2-4}\times\frac{x^2+4x+4}{8x^2}=\frac{4x}{(x-2)(x+2)}\times\frac{(x+2)(x+2)}{8x^2}=\]

Answer 2

Okay, first we'll factor those: x^2 - 4 = (x+2)(x-2) by difference of squares factorization, and x^2 + 4x + 4 = (x+2)(x+2) = (x+2)^2
Then\[{4x \over (x+2)(x-2)} \div {8x^2 \over (x+2)^2} = {4x \over (x+2)(x-2)} \times {(x+2)^2 \over 8x^2}\]
Since dividing by a fraction is the same as multiplying by its reciprocal.
We cancel a factor of 4, x, and (x+2) to get\[{1 \over x-2} \times {x+2 \over 2x} = {x+2 \over 2x(x-2)} = {x+2 \over 2x^2 - 4x}\]

Answer 3

\[\frac{x+2}{2x(x-2)}\]

Answer 4

thanks yall

Answer 5

yw