As it is mentioned
a, b and x are integers greater than zero

so a,b,x can be equal, consecutive numbers or any other possible value matching the criteria.

Now we can glance down straight to the answer choice we can eliminate option B, D and E. Option A is the possible contender.

Option B : if x is subtracted from numerator than the value will be < \(\frac{a}{a+b}\)
Option D : we are squaring the value but remember we are also squaring the denominator so the value will be < \(\frac{a}{a+b}\)
Option E: Again if 1` is subtracted it will only reduce the value so it will always be < \(\frac{a}{a+b}\)

Now for option C:: Let dig some value such as all numbers are considered equal, after solving it < \(\frac{a}{a+b}\)

So only option left is A, we can put some value such as a = b =c = x =1, or any integer value greater than 1, it will always be > \(\frac{a}{a+b}\)
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--------ASIDE---------------------------------------
Here's a nice property of fractions: If a, b and k are positive, then (a + k)/(b + k) approaches 1 as k gets bigger.
For example, the fraction (2+11)/(3+11) is closer to 1 than 2/3 is.
Likewise, the fraction (1+7)/(2+7) is closer to 1 than 1/2 is.
-----ONTO THE QUESTION!!!-------------------------------------

Since a and b are positive, we know that a+b > a
So, the fraction a/(a+b) must be less than 1

So, based on the above property, if we add a positive number to numerator and denominator, the resulting fraction will be closer to 1 than the original fraction is.

Check the answer choices. . . .
(A) (a + x)/(a + b + x)
Since x is positive, we know that (a + x)/(a + b + x) will be closer to 1 than a/(a+b) is.
Since a/(a+b) is less than 1, we know that (a + x)/(a + b + x) will be greater than a/(a+b)

Answer: A
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