Bài giải:

$\left( \frac{5x+y}{{{x}^{2}}-5xy}+\frac{5x-y}{{{x}^{2}}+5xy} \right).\frac{{{x}^{2}}-25{{y}^{2}}}{{{x}^{2}}+{{y}^{2}}}$= $\left( \frac{5x+y}{x(x-5y)}+\frac{5x-y}{x(x+5y)} \right).\frac{(x-5y)(x+5y)}{{{x}^{2}}+{{y}^{2}}}$

= $\left( \frac{(5x+y)(x+5y)}{x(x-5y)(x+5y)}+\frac{(5x-y)(x-5y)}{x(x+5y)(x-5y)} \right).\frac{(x-5y)(x+5y)}{{{x}^{2}}+{{y}^{2}}}$

= $\frac{5{{x}^{2}}+26xy+5{{y}^{2}}+5{{x}^{2}}-26xy+5{{y}^{2}}}{x(x-5y)(x+5y)}.\frac{(x-5y)(x+5y)}{{{x}^{2}}+{{y}^{2}}}$

= $\frac{10{{x}^{2}}+10{{y}^{2}}}{x(x-5y)(x+5y)}.\frac{(x-5y)(x+5y)}{{{x}^{2}}+{{y}^{2}}}$

= $\frac{10}{x}$

Vậy a = 10