Ответ:
а) \(\frac{4x}{3(x+1)}-\frac{x-1}{x+1}=\frac{4x-3(x-1)}{3(x+1)}=\frac{x+3}{3(x+1)}\).
б) Так как \(b-a=-(a-b)\), то:
\(\frac{2a}{3(a-b)}+\frac{a-2}{b-a}=\frac{2a-3(a-2)}{3(a-b)}=\frac{6-a}{3(a-b)}\).
в) \(\frac{4-a}{a(a-2)}+\frac{2}{a}=\frac{4-a+2(a-2)}{a(a-2)}=\frac{a}{a(a-2)}=\frac{1}{a-2}\).
г) \(\frac{5x+3y}{2(x+y)}-\frac{7x+4y}{3(x+y)}=\frac{3(5x+3y)-2(7x+4y)}{6(x+y)}=\frac{x+y}{6(x+y)}=\frac{1}{6}\).
д) \(\frac{4b+8a}{3(b-a)}=-\frac{4b+8a}{3(a-b)}\). Поэтому:
\(\frac{5a+3b}{2(a-b)}+\frac{4b+8a}{3(b-a)}=\frac{3(5a+3b)-2(4b+8a)}{6(a-b)}=\frac{b-a}{6(a-b)}=-\frac{1}{6}\).
Ответ: а) \(\frac{x+3}{3(x+1)}\); б) \(\frac{6-a}{3(a-b)}\); в) \(\frac{1}{a-2}\); г) \(\frac{1}{6}\); д) \(-\frac{1}{6}\).
