22) \(\frac{10y}{(x-y)^2}-\frac{8}{y-x}=\)

Так как \(y-x=-(x-y)\), то

\[ -\frac{8}{y-x}=\frac{8}{x-y}.\]

Приведём дроби к общему знаменателю \((x-y)^2\):

\[\frac{10y}{(x-y)^2}+\frac{8}{x-y}=\frac{10y}{(x-y)^2}+\frac{8(x-y)}{(x-y)^2}.\]

\[\frac{10y+8x-8y}{(x-y)^2}=\frac{8x+2y}{(x-y)^2}=\frac{2(4x+y)}{(x-y)^2}.\]

ОДЗ: \(x\ne y\).

Ответ: \(\frac{2(4x+y)}{(x-y)^2}\).