If `x # y = (x + y)/(xy)`, then the value of `P # (q # r)` for every p, q, in N :

Options

A. a. `(pq + qr + r^2)/(p(q + q + r))`
B. b. `(pqr + q + r)/(p(q + r))`
C. c. `(p(qr + q + r))/(pqr)`
D. d. None of these

Answer

To solve the problem step by step, we will first rewrite the operation defined by \( x \# y = \frac{x + y}{xy} \). We need to find the value of \( P \# (Q \# R) \).

### Step 1: Calculate \( Q \# R \)
Using the operation defined, we have:
\[ Q \# R = \frac{Q + R}{QR} \]

### Step 2: Substitute \( Q \# R \) into \( P \# (Q \# R) \)
Now, we substitute \( Q \# R \) into the expression for \( P \# (Q \# R) \):
\[ P \# (Q \# R) = P \# \left( \frac{Q + R}{QR} \right) \]

### Step 3: Apply the operation \( P \# \left( \frac{Q + R}{QR} \right) \)
Using the definition of the operation again:
\[ P \# \left( \frac{Q + R}{QR} \right) = \frac{P + \frac{Q + R}{QR}}{P \cdot \frac{Q + R}{QR}} \]

### Step 4: Simplify the numerator
The numerator becomes:
\[ P + \frac{Q + R}{QR} = \frac{P \cdot QR + (Q + R)}{QR} \]
So, the numerator is:
\[ \frac{PQR + Q + R}{QR} \]

### Step 5: Simplify the denominator
The denominator is:
\[ P \cdot \frac{Q + R}{QR} = \frac{P(Q + R)}{QR} \]

### Step 6: Combine the results
Now, we can substitute the simplified numerator and denominator back into the expression:
\[ P \# (Q \# R) = \frac{\frac{PQR + Q + R}{QR}}{\frac{P(Q + R)}{QR}} \]

### Step 7: Cancel out the common terms
The \( QR \) in the numerator and denominator cancels out:
\[ P \# (Q \# R) = \frac{PQR + Q + R}{P(Q + R)} \]

### Final Result
Thus, the final value of \( P \# (Q \# R) \) is:
\[ P \# (Q \# R) = \frac{PQR + Q + R}{P(Q + R)} \]