Let `x_(1),x_(2) , x_(3),..........x_(2014)` be real numbers different from 1 such that `x_(1) + x_(2) + .......+x_(2014) =1` and `x_(1)/(1-x_(1))+x_(2)/(1-x_(2))+.......+ x_(2014)/(1-x_(2014))= 1` then `x_(1)^(2)/(1-x_(1)) + x_(2)^(2)/(1-x_(2)) +................+ x_(2014)^(2)/(1-x_(2014))=?`

Answer

To solve the problem, we need to find the value of

\[ S = \frac{x_1^2}{1-x_1} + \frac{x_2^2}{1-x_2} + \ldots + \frac{x_{2014}^2}{1-x_{2014}}. \]

Given the conditions:

1. \( x_1 + x_2 + \ldots + x_{2014} = 1 \) (Equation 1)
2. \( \frac{x_1}{1-x_1} + \frac{x_2}{1-x_2} + \ldots + \frac{x_{2014}}{1-x_{2014}} = 1 \) (Equation 2)

### Step 1: Rewrite Equation 2

From Equation 2, we can rewrite it as:

\[ \sum_{i=1}^{2014} \frac{x_i}{1-x_i} = 1. \]

### Step 2: Subtract Equation 1 from Equation 2

Now, we subtract Equation 1 from Equation 2:

\[ \sum_{i=1}^{2014} \left( \frac{x_i}{1-x_i} - x_i \right) = 0. \]

This simplifies to:

\[ \sum_{i=1}^{2014} \left( \frac{x_i - x_i(1-x_i)}{1-x_i} \right) = 0. \]

### Step 3: Simplify the expression

The expression inside the summation becomes:

\[ \frac{x_i - x_i + x_i^2}{1-x_i} = \frac{x_i^2}{1-x_i}. \]

Thus, we have:

\[ \sum_{i=1}^{2014} \frac{x_i^2}{1-x_i} = 0. \]

### Step 4: Conclusion

Since the sum of the squares divided by \(1-x_i\) equals zero, we conclude that:

\[ S = \frac{x_1^2}{1-x_1} + \frac{x_2^2}{1-x_2} + \ldots + \frac{x_{2014}^2}{1-x_{2014}} = 0. \]

### Final Answer

The value of

\[ \frac{x_1^2}{1-x_1} + \frac{x_2^2}{1-x_2} + \ldots + \frac{x_{2014}^2}{1-x_{2014}} = 0. \]

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