Does the fractional equation $\frac{1}{x-5} +\frac{1}{x+5}=\frac{2x+1}{x^2-25}$ have any solutions? We have a partial fraction equation:
$$\frac{1}{x-5} +\frac{1}{x+5}=\frac{2x+1}{x^2-25}$$

I multiplied the equation by the common denominator $(x+5)(x-5)$ and got $0=1$. Is this correct?
Yes it is equivalent to

$$\frac{2x}{x^2-25}=\frac{2x+1}{x^2-25}\iff 2x=2x+1$$

which indeed has not solutions for $x\in \mathbb R$.
Rearrange the equation as 

$$\frac{2x}{x^2-25}-\frac{1}{x-5}-\frac{1}{x+5} 
=\frac{2x+1}{x^2-25}-\frac{2x}{x^2-25}= \frac{1}{x^2-25}=0$$

which leads to the solutions 

$$x=\pm \infty$$