`P(A, B)=AxxB,S(A,B)=A-B,D(A,B)=A+B`
`P(S(D(7, 3), P(2, 3)), P(6, 5)))` is equal to:

Options

A. 120
B. 60
C. 30
D. none of these

Answer

To solve the expression \( P(S(D(7, 3), P(2, 3)), P(6, 5)) \), we will break it down step by step using the definitions provided:

1. **Calculate \( D(7, 3) \)**:
- According to the definition \( D(A, B) = A + B \).
- Substitute \( A = 7 \) and \( B = 3 \):
\[ D(7, 3) = 7 + 3 = 10 \]

2. **Calculate \( P(2, 3) \)**:
- According to the definition \( P(A, B) = A \times B \).
- Substitute \( A = 2 \) and \( B = 3 \):
\[ P(2, 3) = 2 \times 3 = 6 \]

3. **Calculate \( S(D(7, 3), P(2, 3)) \)**:
- We already found \( D(7, 3) = 10 \) and \( P(2, 3) = 6 \).
- According to the definition \( S(A, B) = A - B \):
\[ S(10, 6) = 10 - 6 = 4 \]

4. **Calculate \( P(6, 5) \)**:
- Again using the definition \( P(A, B) = A \times B \):
- Substitute \( A = 6 \) and \( B = 5 \):
\[ P(6, 5) = 6 \times 5 = 30 \]

5. **Calculate \( P(S(D(7, 3), P(2, 3)), P(6, 5)) \)**:
- We found \( S(D(7, 3), P(2, 3)) = 4 \) and \( P(6, 5) = 30 \).
- Using the definition \( P(A, B) = A \times B \):
\[ P(4, 30) = 4 \times 30 = 120 \]

Thus, the final answer is:
\[ \boxed{120} \]