Question

(-x-¹y)°

Answer 1

$a^0=1\ \text{for any real number except 0}.\\\\\text{Therefore}\ (-x^{-1}y)^0=1\ \text{for}\ x\neq0\ \text{and}\ y\neq0.\\\\a^{-1}=\dfrac{1}{a^1}\ \text{for any real number except 0}.\\\\\text{Therefore}\ w^{-1}=\dfrac{1}{w}.\\\\\dfrac{(-x^{-1}y)^0}{4w^{-1}y^2}=\dfrac{1}{4y^2\cdot\frac{1}{w}}=\dfrac{1}{4y^2}\cdot\dfrac{w}{1}=\boxed{\dfrac{w}{4y^2}}$