Ta có \(\frac{1}{y^2-xy}+\frac{1}{x^2-xy}\)=\(\frac{1}{y\left(y-x\right)}+\frac{1}{x\left(x-y\right)}=\frac{-1}{y\left(x-y\right)}+\frac{1}{x\left(x-y\right)}\)
=\(-\frac{x}{xy\left(x-y\right)}+\frac{y}{xy\left(x-y\right)}\)=\(\frac{-\left(x-y\right)}{xy\left(x-y\right)}\)
=\(-\frac{1}{xy}\) (*)
Thay xy=-1 vào (*)
có -\(\frac{1}{xy}=-\frac{1}{-1}=1\)
Vậy P=1