Question

Divide.

Answer 1

the x^2 + 2x + 1 is thw square of (x + 1)^2
as per the previous questions you can solve it

Answer 2

The first part would be (x -2) 2x? + 1

Answer 3

Yes!

Answer 4

@theredhead1617 do you know how to solve the problem

Answer 5

\[\frac{ \frac{ (x+1)(x+1) }{ x-2 } }{ \frac{ (x-1)(x+1) }{ (x-2)(x+2) } }\]
\[\frac{ (x+1)(x+1) }{ (x-2) } \div { \frac{ (x-1)(x+1) }{ (x-2)(x+2) } }\\]

\frac{ (x+1)(x+1) }{ (x-2) } \times { \frac{ (x-2)(x+2) }{ (x-1)(x+1) } }\\]\]
can you solv further?

Answer 6

No @miteshchvm I dont understand how to. I wish to know where to begin but i do not

Answer 7

can you factorise x^2 + 2x + 1?

Answer 8

x^2 + 2x + 1 = (x + 1)^2 = (x+1)(x+1)

x^2 - 1 = (x +1)(x - 1)

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

you get it?

Answer 9

(x + 1) to second power?

Answer 10

yes

Answer 11

\[x^2 + 2x + 1 = (x + 1)^2 = (x+1)(x+1) \]
\[x^2 - 1 = (x +1)(x - 1) \]
\[x^2 - 4 = (x-2)(x+2)\]
you get it?
i wrote this instead of your question in order to simplify,

Answer 12

just a little bit, its hard

Answer 13

now refer to my first comment

Answer 14

on the previous question?

Answer 15

your question can be written as\[{ \frac{ (x^2 + 2x + 1) }{ x-2 } } \div { \frac{ x^2 - 1 }{ x^2 - 4 } } \]
now
\[\frac{ (x+1)(x+1) }{ (x-2) } \div { \frac{ (x-1)(x+1) }{ (x-2)(x+2) } }\\] \]
\[\frac{ (x+1)(x+1) }{ (x-2) } \times { \frac{ (x-2)(x+2) }{ (x-1)(x+1) } }\\] \]

Answer 16

you get it?

Answer 17

im getting a better understand of it

Answer 18

A:

Answer 19

B:

Answer 20

C:

Answer 21

D:

Answer 22

on eliminating values you get
\[\frac{ (x+1)(x+2) }{ (x-1) }\]

Answer 23

thanks for everything @miteshchvm