Question

(x^2-5x)/(x^2-25)

Answer 1

multiply by the reciprocal

Answer 2

\[\frac{ x^2-5x }{ 1 }\times \frac{ 1 }{ x^2-25 }\]

Answer 3

which =\[\frac{ x^2-5x }{ x^2-25 }\]

Answer 4

\[\frac{ x }{ 5 }\]

Answer 5

Does this help @french_dame ?

Answer 6

yes thank you

Answer 7

your welcome

Answer 8

but the answer it says is x/x+5

Answer 9

hmm.....

Answer 10

@ganeshie8 can you help us?

Answer 11

@phi

Answer 12

@dan815

Answer 13

\[ \frac{x^2-5x}{x^2-25} \]

on the top, notice that there is an "x" in both terms. Can you factor an x out from each term? what do you get?

Answer 14

factoring an x is the opposite of "distributing"

example: x(x-5)= x*x - 5*x or x^2 - 5x

take an x out of each term, and put it "in front"

Answer 15

x(x-5)

Answer 16

now, the bottom
x^2 - 25

this is the ubiquitous "difference of squares"
if you see the pattern
a^2 - b^2
then you know it factors into (a-b)(a+b)

can you use that pattern to factor (x^2 - 25) ?
(btw 25 is 5*5 = 5^2 )

Answer 17

(x-5)(x+5)
right?

Answer 18

oh got it then you cross out the two (x-5)
and end up wth x/(x=5)

Answer 19

(x+5)*

Answer 20

yes.
so far we have
\[ \frac{x(x-5)}{(x-5)(x+5)} \]

anything divided by itself (except zero) is 1
in other words, (x-5)/(x-5) = 1 (as long as x≠5)

Answer 21

thank you

Answer 22

yw