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If x ≠ –1, then 2x^2 + 4x + 2/(x + 1)^2 =

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If x ≠ –1, then 2x^2 + 4x + 2/(x + 1)^2 =  [#permalink]

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New post 09 Apr 2019, 03:47
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

95% (00:47) correct 5% (02:12) wrong based on 19 sessions

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Re: If x ≠ –1, then 2x^2 + 4x + 2/(x + 1)^2 =  [#permalink]

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New post 09 Apr 2019, 03:51
Bunuel wrote:
If x ≠ –1, then \(\frac{2x^2 + 4x + 2}{(x + 1)^2} =\)

(A) 0
(B) 1
(C) 2
(D) 4
(E) 6



\(\frac{2x^2 + 4x + 2}{(x + 1)^2} =\)
2* (x+1)^2/(x+1)^2 = 2
IMO C
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Re: If x ≠ –1, then 2x^2 + 4x + 2/(x + 1)^2 =  [#permalink]

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New post 09 Apr 2019, 04:07
1
Bunuel wrote:
If x ≠ –1, then \(\frac{2x^2 + 4x + 2}{(x + 1)^2} =\)

(A) 0
(B) 1
(C) 2
(D) 4
(E) 6


\(\frac{2x^2 + 4x + 2}{(x + 1)^2} = \frac{2(x^2 + 2x + 1)}{(x + 1)^2} = \frac{2(x + 1)^2}{(x + 1)^2}\) because \((x + 1)^2 = x^2 + 2x + 1\) = 2(Option C)
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Re: If x ≠ –1, then 2x^2 + 4x + 2/(x + 1)^2 =  [#permalink]

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New post 10 Apr 2019, 18:02
Bunuel wrote:
If x ≠ –1, then \(\frac{2x^2 + 4x + 2}{(x + 1)^2} =\)

(A) 0
(B) 1
(C) 2
(D) 4
(E) 6


Simplifying, we have:

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

Answer: C
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Re: If x ≠ –1, then 2x^2 + 4x + 2/(x + 1)^2 =   [#permalink] 10 Apr 2019, 18:02
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