If `Sigmaxy =1 `, then `Sigma (x+y)/( 1-xy)=`

Options

A. `(1)/( xyz)`
B. `(4)/( xyz)`
C. `xyz`
D. none

Answer

To solve the problem, we need to find the value of \(\Sigma \frac{x+y}{1-xy}\) given that \(\Sigma xy = 1\). Let's break this down step by step.

### Step 1: Understand the given information
We are given that \(\Sigma xy = 1\). This means that the sum of the products of the variables \(x\), \(y\), and \(z\) taken two at a time is equal to 1. In mathematical terms, this can be expressed as:
\[ xy + yz + zx = 1 \]

### Step 2: Express the desired summation
We need to find:
\[ \Sigma \frac{x+y}{1-xy} \]
This can be expanded as:
\[ \frac{x+y}{1-xy} + \frac{y+z}{1-yz} + \frac{z+x}{1-zx} \]

### Step 3: Simplify each term
Let's simplify each term in the summation:
1. For the first term \(\frac{x+y}{1-xy}\):
- We can express \(y+z\) as \(1 - xz\) (from the equation \(xy + yz + zx = 1\)).
- Thus, we can rewrite:
\[ \frac{x+y}{1-xy} = \frac{x + (1 - zx)}{1 - xy} = \frac{1 - zx + x}{1 - xy} \]

2. For the second term \(\frac{y+z}{1-yz}\):
- Similarly, we can express \(x+z\) as \(1 - xy\).
- Thus:
\[ \frac{y+z}{1-yz} = \frac{y + (1 - xy)}{1 - yz} = \frac{1 - xy + y}{1 - yz} \]

3. For the third term \(\frac{z+x}{1-zx}\):
- We can express \(x+y\) as \(1 - yz\).
- Thus:
\[ \frac{z+x}{1-zx} = \frac{z + (1 - yz)}{1 - zx} = \frac{1 - yz + z}{1 - zx} \]

### Step 4: Combine the simplified terms
Now we can combine these simplified terms:
\[ \frac{x+y}{1-xy} + \frac{y+z}{1-yz} + \frac{z+x}{1-zx} = \frac{1}{1-xy} + \frac{1}{1-yz} + \frac{1}{1-zx} \]

### Step 5: Find the common denominator
To combine these fractions, we need a common denominator:
\[ \text{Common Denominator} = (1-xy)(1-yz)(1-zx) \]

### Step 6: Evaluate the summation
Using the fact that \(\Sigma xy = 1\), we can substitute and simplify:
\[ \Sigma \frac{x+y}{1-xy} = \frac{1}{z} + \frac{1}{x} + \frac{1}{y} \]
This leads us to:
\[ \Sigma \frac{x+y}{1-xy} = \frac{1}{xyz} \]

### Final Result
Thus, the final result is:
\[ \Sigma \frac{x+y}{1-xy} = \frac{1}{xyz} \]