We can solve this question using matching operations

Given:
Quantity A: \(\frac{a-b}{a+b}\)
Quantity B: \(\frac{a^2-b^2}{a^2+b^2}\)

Factor quantity B to get:
Quantity A: \(\frac{a-b}{a+b}\)
Quantity B: \(\frac{(a+b)(a-b)}{a^2+b^2}\)

Since 0 < b < a, we know that a+b is POSITIVE. So, we can safely multiply both quantities by (a+b) to get:
Quantity A: \(a-b\)
Quantity B: \(\frac{(a+b)(a+b)(a-b)}{a^2+b^2}\)

Since 0 < b < a, we know that a-b is POSITIVE. So, we can safely divide both quantities by (a-b) to get:
Quantity A: \(\frac{a-b}{a-b}\)
Quantity B: \(\frac{(a+b)(a+b)}{a^2+b^2}\)

Simplify to get:
Quantity A: \(1\)
Quantity B: \(\frac{(a+b)(a+b)}{a^2+b^2}\)

Expand Quantity B to get:
Quantity A: \(1\)
Quantity B: \(\frac{a^2 + 2ab + b^2}{a^2+b^2}\)

Rewrite Quantity B to get:
Quantity A: \(1\)
Quantity B: \(\frac{(a^2 + b^2) + 2ab}{a^2+b^2}\)

Useful rule: \(\frac{x+y}{z} = \frac{x}{z} + \frac{y}{z}\)

Use the above rule to rewrite Quantity B as follows:
Quantity A: \(1\)
Quantity B: \(\frac{a^2+b^2}{a^2+b^2}+ \frac{2ab}{a^2+b^2}\)

Simplify Quantity B:
Quantity A: \(1\)
Quantity B: \(1+\frac{2ab}{a^2+b^2}\)

Subtract 1 from both quantities:
Quantity A: \(0\)
Quantity B: \(\frac{2ab}{a^2+b^2}\)

Since a and b are both positive, we know that 2ab is POSITIVE
Likewise, Since a and b are both positive, we know that a² + b² is POSITIVE

We get:
Quantity A: \(0\)
Quantity B: \(\frac{POSITIVE}{POSITIVE}\)

Simplify:
Quantity A: \(0\)
Quantity B: \(POSITIVE\)

Answer: B


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