If `xy + yz + zx = 1` then `sum (x + y)/(1 - xy)=`

Options

A. `(4)/(xyz)`
B. `(1)/(xyz)`
C. xyz
D. `x+ y+ z`

Answer

To solve the problem, we need to find the value of the expression \(\sum \frac{x+y}{1-xy}\) given that \(xy + yz + zx = 1\).

### Step-by-Step Solution:

1. **Start with the given equation**: We have \(xy + yz + zx = 1\).

2. **Rewrite the expression**: We want to evaluate \(\sum \frac{x+y}{1-xy}\). We can express this as:
\[ \sum \frac{x+y}{1-xy} = \sum \frac{x+y}{(xy + yz + zx) - xy} \]
Substituting \(xy + yz + zx = 1\) into the equation gives:
\[ \sum \frac{x+y}{1 - xy} = \sum \frac{x+y}{1 - xy} \]

3. **Simplify the denominator**: The denominator can be rewritten as:
\[ 1 - xy = yz + zx \]
Thus, we have:
\[ \sum \frac{x+y}{1 - xy} = \sum \frac{x+y}{yz + zx} \]

4. **Break down the sum**: The expression can be separated into three parts:
\[ \frac{x+y}{yz+zx} + \frac{y+z}{zx+xy} + \frac{z+x}{xy+yz} \]

5. **Factor out common terms**: Notice that each term can be manipulated:
\[ \frac{x+y}{yz+zx} = \frac{x+y}{z(y+x)} \]
This leads to:
\[ = \frac{1}{z} \quad \text{(for each term)} \]

6. **Sum up the fractions**: Therefore, we can write:
\[ \sum \frac{x+y}{1 - xy} = \frac{1}{z} + \frac{1}{x} + \frac{1}{y} \]

7. **Combine the fractions**: The final step is to combine these fractions:
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{xy + xz + yz}{xyz} \]
Since \(xy + yz + zx = 1\), we can substitute this in:
\[ = \frac{1}{xyz} \]

8. **Final result**: Therefore, the value of the expression \(\sum \frac{x+y}{1-xy}\) is:
\[ \frac{1}{xyz} \]

### Conclusion:
The final answer is \(\frac{1}{xyz}\).