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chicagochica5
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miteshchvm
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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
chicagochica5
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The first part would be (x -2) 2x? + 1
theredhead1617
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Yes!
chicagochica5
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@theredhead1617 do you know how to solve the problem
miteshchvm
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\[\frac{ \frac{ (x+1)(x+1) }{ x-2 } }{ \frac{ (x-1)(x+1) }{ (x-2)(x+2) } }\]
\[\frac{ (x+1)(x+1) }{ (x-2) } \div { \frac{ (x-1)(x+1) }{ (x-2)(x+2) } }\\]
\frac{ (x+1)(x+1) }{ (x-2) } \times { \frac{ (x-2)(x+2) }{ (x-1)(x+1) } }\\]\]
can you solv further?
chicagochica5
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No @miteshchvm I dont understand how to. I wish to know where to begin but i do not
miteshchvm
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can you factorise x^2 + 2x + 1?
miteshchvm
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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?
chicagochica5
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(x + 1) to second power?
miteshchvm
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yes
miteshchvm
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\[x^2 + 2x + 1 = (x + 1)^2 = (x+1)(x+1) \]
\[x^2 - 1 = (x +1)(x - 1) \]
\[x^2 - 4 = (x-2)(x+2)\]
you get it?
i wrote this instead of your question in order to simplify,
chicagochica5
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just a little bit, its hard
miteshchvm
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now refer to my first comment
chicagochica5
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on the previous question?
miteshchvm
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your question can be written as\[{ \frac{ (x^2 + 2x + 1) }{ x-2 } } \div { \frac{ x^2 - 1 }{ x^2 - 4 } } \]
now
\[\frac{ (x+1)(x+1) }{ (x-2) } \div { \frac{ (x-1)(x+1) }{ (x-2)(x+2) } }\\] \]
\[\frac{ (x+1)(x+1) }{ (x-2) } \times { \frac{ (x-2)(x+2) }{ (x-1)(x+1) } }\\] \]
miteshchvm
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you get it?
chicagochica5
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im getting a better understand of it
chicagochica5
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A:
chicagochica5
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B:
chicagochica5
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C:
chicagochica5
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D:
miteshchvm
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on eliminating values you get
\[\frac{ (x+1)(x+2) }{ (x-1) }\]
chicagochica5
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thanks for everything @miteshchvm