Bài giải:

ĐKXĐ: $x\ne 0$; $x\ne -5$

$\frac{{{x}^{2}}+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x(x+5)}$= $\frac{{{x}^{2}}+2x}{2(x+5)}+\frac{x-5}{x}+\frac{50-5x}{2x(x+5)}$

= $\frac{x({{x}^{2}}+2x)}{2x(x+5)}+\frac{2(x-5)(x+5)}{2x(x+5)}+\frac{50-5x}{2x(x+5)}$

=  $\frac{{{x}^{3}}+2{{x}^{2}}+2{{x}^{2}}-50+50-5x}{2x(x+5)}$

= $\frac{{{x}^{3}}+4{{x}^{2}}-5x}{2x(x+5)}$ = $\frac{x(x-1)(x+5)}{2x(x+5)}$ = $\frac{x-1}{2}$

Vậy đáp án đúng là: A