Ответ:

а) \[\frac{a-1}{2(a-4)}+\frac{a}{a-4}=\frac{a-1+2a}{2(a-4)}=\frac{3a-1}{2(a-4)}.\]

б) Так как \(3x-12=3(x-4)\), а \(2x-8=2(x-4)\), получаем: \[\frac{x-1}{3(x-4)}-\frac{x-3}{2(x-4)}=\frac{2(x-1)-3(x-3)}{6(x-4)}=\frac{7-x}{6(x-4)}.\]

в) \(4y-4=4(y-1)\), \(5-5y=-5(y-1)\): \[\frac{3y}{4(y-1)}+\frac{2y}{-5(y-1)}=\frac{15y-8y}{20(y-1)}=\frac{7y}{20(y-1)}.\]

Ответ: а) \(\frac{3a-1}{2(a-4)}\); б) \(\frac{7-x}{6(x-4)}\); в) \(\frac{7y}{20(y-1)}\).