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the x^2 + 2x + 1 is thw square of (x + 1)^2 as per the previous questions you can solve it
The first part would be (x -2) 2x? + 1

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Other answers:

Yes!
@theredhead1617 do you know how to solve the problem
\[\frac{ \frac{ (x+1)(x+1) }{ x-2 } }{ \frac{ (x-1)(x+1) }{ (x-2)(x+2) } }\] \[\frac{ (x+1)(x+1) }{ (x-2) } \div { \frac{ (x-1)(x+1) }{ (x-2)(x+2) } }\\] \frac{ (x+1)(x+1) }{ (x-2) } \times { \frac{ (x-2)(x+2) }{ (x-1)(x+1) } }\\]\] can you solv further?
No @miteshchvm I dont understand how to. I wish to know where to begin but i do not
can you factorise x^2 + 2x + 1?
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?
(x + 1) to second power?
yes
\[x^2 + 2x + 1 = (x + 1)^2 = (x+1)(x+1) \] \[x^2 - 1 = (x +1)(x - 1) \] \[x^2 - 4 = (x-2)(x+2)\] you get it? i wrote this instead of your question in order to simplify,
just a little bit, its hard
now refer to my first comment
on the previous question?
your question can be written as\[{ \frac{ (x^2 + 2x + 1) }{ x-2 } } \div { \frac{ x^2 - 1 }{ x^2 - 4 } } \] now \[\frac{ (x+1)(x+1) }{ (x-2) } \div { \frac{ (x-1)(x+1) }{ (x-2)(x+2) } }\\] \] \[\frac{ (x+1)(x+1) }{ (x-2) } \times { \frac{ (x-2)(x+2) }{ (x-1)(x+1) } }\\] \]
you get it?
im getting a better understand of it
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on eliminating values you get \[\frac{ (x+1)(x+2) }{ (x-1) }\]
thanks for everything @miteshchvm

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