4) а) \(\frac{4x}{3(x+1)}-\frac{x-1}{x+1}\); б) \(\frac{2a}{3(a-b)}+\frac{a-2}{b-a}\); в) \(\frac{4-a}{a(a-2)}+\frac{2}{a}\).

а) Приводим к знаменателю \(3(x+1)\):

\(\frac{4x}{3(x+1)}-\frac{x-1}{x+1}=\frac{4x-3(x-1)}{3(x+1)}=\frac{4x-3x+3}{3(x+1)}=\frac{x+3}{3(x+1)}\), \(x\ne-1\).

б) Так как \(b-a=-(a-b)\), получаем:

\(\frac{2a}{3(a-b)}+\frac{a-2}{b-a}=\frac{2a}{3(a-b)}-\frac{a-2}{a-b}=\frac{2a-3(a-2)}{3(a-b)}=\frac{6-a}{3(a-b)}\), \(a\ne b\).

в) Приводим к знаменателю \(a(a-2)\):

\(\frac{4-a}{a(a-2)}+\frac{2}{a}=\frac{4-a+2(a-2)}{a(a-2)}=\frac{a}{a(a-2)}=\frac{1}{a-2}\), \(a\ne0\), \(a\ne2\).

Ответ: а) \(\frac{x+3}{3(x+1)}\); б) \(\frac{6-a}{3(a-b)}\); в) \(\frac{1}{a-2}\).