c, x+5/x^2-5x-x+25/2x^2-50=x-5/2x^2+10x

c, ĐKXĐ:

$\left\{\begin{array}{l}{x}^2-5x\ne 0\\ 2x^2-50\ne 0\\ 2x^2+10x\end{array}\right.\Rightarrow x\ne 0; x\ne 0\pm 5$

Ta có:

$$\begin{gathered} \frac{{\displaystyle x+5}}{x^2-5x}-\frac{{\displaystyle x+25}}{2x^2-50}=\frac{{\displaystyle x-5}}{2x^2+10x} \\ \iff \frac{2{\left(x+5\right)}^2-\left(2x^2+50\right)}{x\left(2x^2-50\right)}=\frac{{\displaystyle {\left(x-5\right)}^2}}{x\left(2x^2-50\right)} \end{gathered}$$

⇔ 2( x² + 10x + 25 ) - ( x² + 25x ) = x² - 10x + 25

⇔ x² - 5x + 50 = x² - 10x + 25

⇔ 5x = - 25 ⇔ x = - 5.

Vậy phương trình đã cho có nghiệm x = - 5.