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2/3 < 1, so sq rt(2/3) > 1.
D and E are less by some calculations.
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15 sec.
A
eliminated D and E as there are negatives
2^2/3^2 < 2/3 < [2][/2]
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out of a/b/c a is surely greatest because diff between numerator and denominator of each term is least in A.
d and e can be calculated to see that d is less than 1 and e is less than 2
A is greater than 2 since each term in A is greater than 0.5.
Hence answer - A
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Which of the following represents the greatest value?

A) \(\sqrt{2} / \sqrt{3} + \sqrt{3}/\sqrt{4}+\sqrt{4}/\sqrt{5}+\sqrt{5}/\sqrt{6}\)

B) \(2/3 + 3/4 + 4/5 + 5/6\)

C) \(2^2/3^2 + 3^2/4^2 + 4^2/5^2 + 5^2/6^2\)

D) \(1-1/3 + 4/5 - 3/4\)

E) \(1-3/4 + 4/5 + 1/3\)

We compared A B C and apply the following rule:

If 0 < k < 1, then 0 < k^2 < k < SQRT k <1

But before that, we need to compare B & D & E.
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√2/√3 vs 2/3

√2/√3 - Multiply num and denom by √3
(√2 * √3) /3 vs 2/3.
3 is equal.

(√2 * √3) VS (√2 * √2)

√2 and √3 are both >1. So (√2 * √3) > (√2 * √2). Similarly for all options.
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