Решение:

1. а) $$\frac{5x-7}{6x} - \frac{x-3}{6x} + \frac{2x-8}{6x} = \frac{5x-7-(x-3)+2x-8}{6x} = \frac{5x-7-x+3+2x-8}{6x} = \frac{6x-12}{6x} = \frac{6(x-2)}{6x} = \frac{x-2}{x}$$
2. б) $$\frac{8y-5}{7y} - \frac{2y-1}{7y} - \frac{10-y}{7y} = \frac{8y-5-(2y-1)-(10-y)}{7y} = \frac{8y-5-2y+1-10+y}{7y} = \frac{7y-14}{7y} = \frac{7(y-2)}{7y} = \frac{y-2}{y}$$
3. в) $$\frac{x-5}{x^2-49} + \frac{12}{x^2-49} = \frac{x-5+12}{x^2-49} = \frac{x+7}{x^2-49} = \frac{x+7}{(x-7)(x+7)} = \frac{1}{x-7}$$
4. г) $$\frac{y^2+2y}{y^2-4y+4} - \frac{4y}{y^2-4y+4} = \frac{y^2+2y-4y}{y^2-4y+4} = \frac{y^2-2y}{y^2-4y+4} = \frac{y(y-2)}{(y-2)^2} = \frac{y}{y-2}$$
5. д) $$\frac{3z}{z^2-2z} - \frac{8-z}{z^2-2z} = \frac{3z-(8-z)}{z^2-2z} = \frac{3z-8+z}{z^2-2z} = \frac{4z-8}{z^2-2z} = \frac{4(z-2)}{z(z-2)} = \frac{4}{z}$$