Question 1209787
Given that $5^x = 7^y = 1225$, we can use the following steps to find $\frac{xy}{x+y}$.

1. Express 1225 as a product of its prime factors.
2. Substitute the prime factorization of 1225 into the given equations.
3. Express x and y in terms of the exponents of the prime factors.
4. Substitute the expressions for x and y into the expression $\frac{xy}{x+y}$.
5. Simplify the expression.

Here are the detailed steps:

1. Express 1225 as a product of its prime factors.
   
   $1225 = 5^2 \cdot 7^2$

2. Substitute the prime factorization of 1225 into the given equations.

   $5^x = 5^2 \cdot 7^2$
   $7^y = 5^2 \cdot 7^2$

3. Express x and y in terms of the exponents of the prime factors.

   $x = 2 + 2 \log_5 7$
   $y = 2 + 2 \log_7 5$

4. Substitute the expressions for x and y into the expression $\frac{xy}{x+y}$.

   $\frac{xy}{x+y} = \frac{(2 + 2 \log_5 7)(2 + 2 \log_7 5)}{(2 + 2 \log_5 7) + (2 + 2 \log_7 5)}$

5. Simplify the expression.

   $\frac{xy}{x+y} = \frac{4 + 4 \log_5 7 + 4 \log_7 5 + 4 \log_5 7 \log_7 5}{4 + 2 \log_5 7 + 2 \log_7 5}$
   
   $\frac{xy}{x+y} = \frac{2 + 2 \log_5 7 + 2 \log_7 5 + 2 \log_5 7 \log_7 5}{2 + \log_5 7 + \log_7 5}$
   
   $\frac{xy}{x+y} = \frac{2 (1 + \log_5 7 + \log_7 5 + \log_5 7 \log_7 5)}{2 + \log_5 7 + \log_7 5}$
   
   $\frac{xy}{x+y} = \frac{2 (1 + \log_5 7)(1 + \log_7 5)}{2 + \log_5 7 + \log_7 5}$

   Since $\log_a b = \frac{1}{\log_b a}$, we have $\log_5 7 \log_7 5 = 1$.  Therefore,
   \[\frac{xy}{x + y} = \frac{2 (1 + \log_5 7)(1 + \log_7 5)}{2 + \log_5 7 + \log_7 5} = \frac{2 (1 + \log_5 7 + \log_7 5 + 1)}{2 + \log_5 7 + \log_7 5} = \boxed{2}.\]