The expression `(x+y)^(-1).(x^(-1) + y^(-1))` is equivalent to :

Options

A. 1
B. `(xy)^(-1)`
C. `x^y`
D. `xy^(-1) + x^(-1)y`

Answer

To solve the expression \((x+y)^{-1} \cdot (x^{-1} + y^{-1})\), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ (x+y)^{-1} \cdot (x^{-1} + y^{-1}) \]
We know that \(x^{-1} = \frac{1}{x}\) and \(y^{-1} = \frac{1}{y}\), so we can rewrite \(x^{-1} + y^{-1}\) as:
\[ x^{-1} + y^{-1} = \frac{1}{x} + \frac{1}{y} \]

### Step 2: Find a common denominator
To add \(\frac{1}{x}\) and \(\frac{1}{y}\), we need a common denominator, which is \(xy\):
\[ \frac{1}{x} + \frac{1}{y} = \frac{y}{xy} + \frac{x}{xy} = \frac{x + y}{xy} \]

### Step 3: Substitute back into the expression
Now, we substitute this back into our original expression:
\[ (x+y)^{-1} \cdot \left(\frac{x + y}{xy}\right) \]

### Step 4: Simplify the expression
Now we can simplify:
\[ (x+y)^{-1} \cdot \frac{x+y}{xy} = \frac{1}{x+y} \cdot \frac{x+y}{xy} \]
The \(x+y\) terms cancel out:
\[ = \frac{1}{xy} \]

### Step 5: Rewrite the final expression
We can rewrite \(\frac{1}{xy}\) as:
\[ xy^{-1} \]

Thus, the expression \((x+y)^{-1} \cdot (x^{-1} + y^{-1})\) is equivalent to:
\[ xy^{-1} \]

### Final Answer
The expression is equivalent to:
\[ xy^{-1} \]

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