ok i will try explain more thoroughly.
For b it is obvious what the result is right?
What i did is trying to get into the "same format" as b as in:
it is pretty obvious that A has similar fractions except they are all square rooted. So to get them into same format multiply each fraction by "1" to clear out the square root. So we do:
sqrt(2)/sqrt(3) *(sqrt(6)/sqrt(6)) ..... + sqrt(5)/sqrt(6) *(sqrt(30)/sqrt(30)). you notice that the sqrt multipliers are all "1s" and that they are the products of the fraction values in question - sqrt(6) = sqrt(2)*sqrt(3) for example.
This way you get something like:
2/3 *(sqrt(3)/sqrt(2)) + .... 5/6 *(sqrt(6)/sqrt(5)).
So you have the basic pattern like B but each fraction has a multiple... you should notice that those multiples (sqrt(3)/sqrt(2)) etc.. result in a value GREATER than 1. so basically each similar fraction like 2/3 is multiplying by something greater than 1. With that you can pretty much deduce that the result will be greater than B's. Hope it helps because there's no doubt you will see something like this on the exam!!!!
If you like my answers please +1 kudos!