Решение:

1. a) $$\frac{3x}{x-4} - \frac{x+8}{x-4} = \frac{3x - (x+8)}{x-4} = \frac{3x - x - 8}{x-4} = \frac{2x - 8}{x-4} = \frac{2(x-4)}{x-4} = 2$$
2. б) $$\frac{y-x}{y^2-9} - \frac{x-3}{9-y^2} = \frac{y-x}{y^2-9} + \frac{x-3}{y^2-9} = \frac{y-x + x - 3}{y^2-9} = \frac{y-3}{y^2-9} = \frac{y-3}{(y-3)(y+3)} = \frac{1}{y+3}$$
3. в) $$\frac{2x+17}{49-x^2} - \frac{10+x}{49-x^2} = \frac{2x+17 - (10+x)}{49-x^2} = \frac{2x+17 - 10 - x}{49-x^2} = \frac{x+7}{49-x^2} = \frac{x+7}{(7-x)(7+x)} = \frac{1}{7-x}$$
4. г) $$\frac{y^2-8y}{y-3} - \frac{9+2y}{3-y} = \frac{y^2-8y}{y-3} + \frac{9+2y}{y-3} = \frac{y^2-8y + 9+2y}{y-3} = \frac{y^2 - 6y + 9}{y-3} = \frac{(y-3)^2}{y-3} = y-3$$
5. д) $$\frac{x^2-6x}{x^2-16} - \frac{2x-16}{x^2-16} = \frac{x^2-6x - (2x-16)}{x^2-16} = \frac{x^2-6x - 2x + 16}{x^2-16} = \frac{x^2 - 8x + 16}{x^2-16} = \frac{(x-4)^2}{(x-4)(x+4)} = \frac{x-4}{x+4}$$
6. е) $$\frac{2x^2+2x}{4x^2-y^2} + \frac{xy+y}{y^2-4x^2} = \frac{2x^2+2x}{4x^2-y^2} - \frac{xy+y}{4x^2-y^2} = \frac{2x^2+2x - (xy+y)}{4x^2-y^2} = \frac{2x^2+2x - xy - y}{4x^2-y^2} = \frac{2x(x+1) - y(x+1)}{(2x-y)(2x+y)} = \frac{(2x-y)(x+1)}{(2x-y)(2x+y)} = \frac{x+1}{2x+y}$$