4) а) \(\frac{x+4}{xy-x^2}+\frac{y+4}{xy-y^2}\); в) \(\frac{4}{c^2-9}-\frac{2}{c^2+3c}\).

а) Разложим знаменатели: \(xy-x^2=x(y-x)\), \(xy-y^2=y(x-y)=-y(y-x)\). Тогда
\[\frac{x+4}{x(y-x)}+\frac{y+4}{y(x-y)}=\frac{x+4}{x(y-x)}-\frac{y+4}{y(y-x)}\]

\[=\frac{y(x+4)-x(y+4)}{xy(y-x)}=\frac{xy+4y-xy-4x}{xy(y-x)}=\frac{4(y-x)}{xy(y-x)}=\frac4{xy}.\]

ОДЗ: \(x\ne0\), \(y\ne0\), \(x\ne y\).

в) \(\frac4{(c-3)(c+3)}-\frac2{c(c+3)}=\frac{4c-2(c-3)}{c(c-3)(c+3)}=\frac{2c+6}{c(c-3)(c+3)}=\frac{2(c+3)}{c(c-3)(c+3)}=\frac2{c(c-3)}.\)

ОДЗ: \(c\ne-3\), \(c\ne0\), \(c\ne3\).

Ответ: а) \(\frac4{xy}\); в) \(\frac2{c(c-3)}\).