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If a > 0, b > 0, and c > 0, a + 1/(b + 1/c) =

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If a > 0, b > 0, and c > 0, a + 1/(b + 1/c) =  [#permalink]

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New post 30 Jul 2018, 00:43
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A
B
C
D
E

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  5% (low)

Question Stats:

92% (01:04) correct 8% (01:24) wrong based on 26 sessions

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If \(a > 0\), \(b > 0\) , and \(c > 0\) , \(a + \frac{1}{b + \frac{1}{c}}\) =


A. \(\frac{a + b}{c}\)

B. \(\frac{ac + bc + 1}{c}\)

C. \(\frac{abc + b + c}{bc}\)

D. \(\frac{a + b + c}{abc + 1}\)

E. \(\frac{abc + a + c}{bc + 1}\)
Spoiler: OA

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If a > 0, b > 0, and c > 0, a + 1/(b + 1/c) =  [#permalink]

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New post 30 Jul 2018, 01:44
Bunuel wrote:
If \(a > 0\), \(b > 0\) , and \(c > 0\) , \(a + \frac{1}{b + \frac{1}{c}}\) =


A. \(\frac{a + b}{c}\)

B. \(\frac{ac + bc + 1}{c}\)

C. \(\frac{abc + b + c}{bc}\)

D. \(\frac{a + b + c}{abc + 1}\)

E. \(\frac{abc + a + c}{bc + 1}\)



It's better to solve this one part to part:

= \(b + \frac{1}{c}\)

=\(\frac{bc + 1}{c}\)

Then

=\(\frac{1}{(bc + 1) / c}\)

= \(\frac{c}{bc + 1}\)

last part :

= a + \(\frac{c}{bc +1}\)

=\(\frac{c + abc + a}{bc +1}\)

The best answer is E.
If a > 0, b > 0, and c > 0, a + 1/(b + 1/c) = &nbs [#permalink] 30 Jul 2018, 01:44
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