Given two operators \( \oplus \) and \( \odot \) on numbers \( p \text{ and } q \) such that \[ p \oplus q = \frac{p^2 + q^2}{pq} \text{and} p \odot q = \frac{p^2}{q}, \] if \( x \oplus y = 2 \odot 2 \), then \( x = \)
We are given the following operations: \[ p \oplus q = \frac{p^2 + q^2}{pq}, p \odot q = \frac{p^2}{q}. \]
Step 1: Calculate \( 2 \odot 2 \).
Using the definition of the \( \odot \) operation, we get: \[ 2 \odot 2 = \frac{2^2}{2} = \frac{4}{2} = 2. \]
Step 2: Solve the equation \( x \oplus y = 2 \).
Substitute into the equation for \( x \oplus y \): \[ x \oplus y = \frac{x^2 + y^2}{xy}. \] We are told that \( x \oplus y = 2 \), so we have: \[ \frac{x^2 + y^2}{xy} = 2. \]
Step 3: Solve for \( x \).
Multiply both sides of the equation by \( xy \): \[ x^2 + y^2 = 2xy. \] Rearranging terms: \[ x^2 - 2xy + y^2 = 0. \] This simplifies to: \[ (x - y)^2 = 0, \] so \( x = y \).
Final Answer: \[ y \]
A function, \( \lambda \), is defined by \[ \lambda ( p,q ) = \begin{cases} (p - q)^2, & \text{if } p \geq q, \\ p + q, & \text{if } p < q. \end{cases} \] The value of the expression \( \dfrac{\lambda ( -(-3 + 2), (-2 + 3) )}{( -(-2 + 1) )} \) is:


