Решение:

a) $$\frac{x}{y-1} + \frac{5}{1-y} = \frac{x}{y-1} - \frac{5}{y-1} = \frac{x-5}{y-1}$$

б) $$\frac{a}{c-3} - \frac{6}{3-c} = \frac{a}{c-3} + \frac{6}{c-3} = \frac{a+6}{c-3}$$

в) $$\frac{2m}{m-n} + \frac{2n}{n-m} = \frac{2m}{m-n} - \frac{2n}{m-n} = \frac{2m-2n}{m-n} = \frac{2(m-n)}{m-n} = 2$$

г) $$\frac{5p}{2q-p} + \frac{10q}{p-2q} = \frac{5p}{2q-p} - \frac{10q}{2q-p} = \frac{5p-10q}{2q-p} = \frac{5(p-2q)}{2q-p} = -5$$

д) $$\frac{a^2+16}{a-4} + \frac{8a}{4-a} = \frac{a^2+16}{a-4} - \frac{8a}{a-4} = \frac{a^2+16-8a}{a-4} = \frac{a^2-8a+16}{a-4} = \frac{(a-4)^2}{a-4} = a-4$$

e) $$\frac{x^2+9y^2}{x-3y} + \frac{6xy}{3y-x} = \frac{x^2+9y^2}{x-3y} - \frac{6xy}{x-3y} = \frac{x^2+9y^2-6xy}{x-3y} = \frac{x^2-6xy+9y^2}{x-3y} = \frac{(x-3y)^2}{x-3y} = x-3y$$