Let `emptyset` be an operation on x and y defined as `x emptysety=(x^(-2)+y^(-2))/(x^(-1)+y^(-1))` . Find the value of `(1emptyset1)emptyset3` ?

Options

A. 0.83
B. `1.00`
C. `2.50`
D. `2.67`

Answer

To solve the problem, we need to evaluate the operation defined as \( x \emptyset y = \frac{x^{-2} + y^{-2}}{x^{-1} + y^{-1}} \) for the expression \( (1 \emptyset 1) \emptyset 3 \).

### Step 1: Calculate \( 1 \emptyset 1 \)

Using the operation definition:
\[ 1 \emptyset 1 = \frac{1^{-2} + 1^{-2}}{1^{-1} + 1^{-1}} \]

Calculating the individual terms:
- \( 1^{-2} = 1 \)
- \( 1^{-1} = 1 \)

Substituting these values into the expression:
\[ 1 \emptyset 1 = \frac{1 + 1}{1 + 1} = \frac{2}{2} = 1 \]

### Step 2: Calculate \( (1 \emptyset 1) \emptyset 3 \)

Now we substitute \( 1 \emptyset 1 \) into the operation with \( 3 \):
\[ 1 \emptyset 3 = \frac{1^{-2} + 3^{-2}}{1^{-1} + 3^{-1}} \]

Calculating the individual terms:
- \( 1^{-2} = 1 \)
- \( 3^{-2} = \frac{1}{9} \)
- \( 1^{-1} = 1 \)
- \( 3^{-1} = \frac{1}{3} \)

Substituting these values into the expression:
\[ 1 \emptyset 3 = \frac{1 + \frac{1}{9}}{1 + \frac{1}{3}} \]

### Step 3: Simplify the numerator and denominator

Calculating the numerator:
\[ 1 + \frac{1}{9} = \frac{9}{9} + \frac{1}{9} = \frac{10}{9} \]

Calculating the denominator:
\[ 1 + \frac{1}{3} = \frac{3}{3} + \frac{1}{3} = \frac{4}{3} \]

### Step 4: Combine the results

Now we can substitute back into our expression:
\[ 1 \emptyset 3 = \frac{\frac{10}{9}}{\frac{4}{3}} = \frac{10}{9} \cdot \frac{3}{4} = \frac{30}{36} = \frac{5}{6} \]

### Final Result

Thus, the value of \( (1 \emptyset 1) \emptyset 3 \) is:
\[ \frac{5}{6} \]