`x"@"y=(x+y)^(2)`
`x#y=(x-y)^(2)`
`(2" @ "2)+(3#3)-(4"@"4)+(5#5)` is equal to :

Options

A. 64
B. `-48`
C. 28
D. none of these

Answer

To solve the expression \((2"@2)+(3#3)-(4"@4)+(5#5)\), we first need to understand the operations defined by the symbols "@" and "#".

1. **Understanding the operations**:
- \(x"@y = (x + y)^2\)
- \(x#y = (x - y)^2\)

2. **Calculating each part of the expression**:
- **Calculate \(2"@2\)**:
\[ 2"@2 = (2 + 2)^2 = 4^2 = 16 \]

- **Calculate \(3#3\)**:
\[ 3#3 = (3 - 3)^2 = 0^2 = 0 \]

- **Calculate \(4"@4\)**:
\[ 4"@4 = (4 + 4)^2 = 8^2 = 64 \]

- **Calculate \(5#5\)**:
\[ 5#5 = (5 - 5)^2 = 0^2 = 0 \]

3. **Putting it all together**:
Now substitute these values back into the original expression:
\[ (2"@2) + (3#3) - (4"@4) + (5#5) = 16 + 0 - 64 + 0 \]

4. **Simplifying the expression**:
\[ 16 + 0 - 64 + 0 = 16 - 64 = -48 \]

Therefore, the final answer is:
\[ \boxed{-48} \]