Diivide and write in lowest terms: x^2-25 over (/) 5+x divided by 5-x over x?

Question

please show me the steps so I can see what I'm doing wrong. thanx for the help.

Answer 1

ok, you've got a difference of squares there on the top. remember that (a^2-b^2) = (a + b)(a-b).

hope this helps; you can probably get it from there :-)

Answer 2

x^2-25 can be factored to (x-5)(x+5) and you divide that by 5+x, when you do that you are left with x-5/1. Next you divide 5-x over x. Turn it into a multiplication and invert your fraction to read x/5-x. Your problem should now read x-5/1 * x/5-x. Good luck!

Answer 3

Start by looking at the equation:

(x^2 - 25) / (5 + x) divided by (5-x) / x
First, factor the numerator of the first term:
(x - 5)(x+5) / (5+x) divided by (5-x)/x
Cancel the (x+5) in the first term:
(x-5) divided by (5-x)/x
Remember that dividing by a fraction is the same as multiplying by its reciprocal:
(x - 5) * x/(5-x)
Multiply the second term by -1/-1
(x-5) * -x/(x-5)
Cancel the (x-5):
-x
is the answer.

Answer 4

x^2-25 = (x-5)*(x+5)
if you divide that by 5+x you get (x-5),since (x+5)/(x+5)=1
Then, you are left with (x-5)/(5-x)/x.
5-x = -1 *(x-5), thus (x-5)/(5-x) = -1
The answer is -1/x

Answer 5

you need to reword this question so its better to understand........ and then maybe i can help