Решение:
- \(\frac{16}{x-4} - \frac{x^2}{x-4} = \frac{16-x^2}{x-4} = \frac{(4-x)(4+x)}{x-4} = \frac{-(x-4)(4+x)}{x-4} = -(4+x)\)
- \(\frac{25}{a+5} - \frac{a^2}{a+5} = \frac{25-a^2}{a+5} = \frac{(5-a)(5+a)}{a+5} = 5-a\)
- \(\frac{3a-1}{a^2-b^2} - \frac{3b-1}{a^2-b^2} = \frac{3a-1-(3b-1)}{a^2-b^2} = \frac{3a-1-3b+1}{a^2-b^2} = \frac{3a-3b}{a^2-b^2} = \frac{3(a-b)}{(a-b)(a+b)} = \frac{3}{a+b}\)
- \(\frac{x-3}{x^2-64} + \frac{11}{x^2-64} = \frac{x-3+11}{x^2-64} = \frac{x+8}{x^2-64} = \frac{x+8}{(x-8)(x+8)} = \frac{1}{x-8}\)
- \(\frac{2a+b}{(a-b)^2} + \frac{a+2b}{(a-b)^2} = \frac{2a+b+a+2b}{(a-b)^2} = \frac{3a+3b}{(a-b)^2} = \frac{3(a+b)}{(a-b)^2}\)
- \(\frac{13x+6y}{(x+y)^2}\)
Ответ: а) -(x+4); б) 5-a; в) 3/(a+b); г) 1/(x-8); д) 3(a+b)/(a-b)^2; е) (13x+6y)/(x+y)^2.