Упростите. заранее спасибо а)х+5/5х-25 - 3х+5/х²-25 б)у-8/4у-16 + 4/у²-4у

Решите задачу:

$\frac{x+5}{5x-25} -\frac{3x+5}{x^2-25} =\frac{x+5}{5(x-5)} -\frac{3x+5}{(x-5)(x+5)} = \frac{(x+5)(x+5)}{5(x-5)(x+5)} -\frac{5(3x+5)}{5(x-5)(x+5)} =\\ =\frac{(x+5)^2-5(3x+5)}{5(x^2-25)} =\frac{x^2+10x+25-15x-25}{5(x^2-25)} =\frac{x^2-5x}{5(x^2-25)}=\\=\frac{x(x-5)}{5(x-5)(x+5)} = \frac{x}{5(x+5)}$

$\frac{y-8}{4y-16} +\frac{4}{y^2-4y} =\frac{y-8}{4(y-4)} +\frac{4}{y(y-4)}= \frac{y(y-8)}{4y(y-4)} +\frac{4*4}{4y(y-4)}= \\ = \frac{y(y-8)+4*4}{4y(y-4)} =\frac{y^2-8y+16}{4y(y-4)}= \frac{(y-4)^2}{4y(y-4)} =\frac{y-4}{4y}$