\(\frac{-8}{15}\)
\(\frac{8}{15}\)
\(\frac{17}{15}\)
\(\frac{-17}{15}\)
Correct answer is A
p>\(x \ast y = \frac{x^2 - y^2}{2xy}\\
=\frac{(x+y)(x-y)}{2xy}\\
-5 \ast 3 =\frac{(-5+3)(-5-3)}{2(-5\times3)}\\
=\frac{-2 \times -8}{2(-5\times3)}\\
=\frac{-8}{15}\)
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