\(A=\dfrac{x}{5-x}+\left(\dfrac{x}{x^2-25}+\dfrac{5-x}{5x+x^2}\right):\dfrac{2x-5}{x^2+5x}=\dfrac{x}{5-x}+\left[\dfrac{x^2}{x\left(x-5\right)\left(x+5\right)}-\dfrac{\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}\right].\dfrac{x^2+5x}{2x-5}=\dfrac{x}{5-x}+\dfrac{x^2-\left(x-5\right)^2}{x\left(x-5\right)\left(x+5\right)}.\dfrac{x\left(x+5\right)}{2x-5}=\dfrac{x}{5-x}+\dfrac{\left(x^2-x^2+10x-25\right)x\left(x+5\right)}{x\left(x-5\right)\left(x+5\right)\left(2x-5\right)}=\dfrac{x}{5-x}+\dfrac{5\left(2x-5\right)x\left(x+5\right)}{x\left(x-5\right)\left(x+5\right)\left(2x-5\right)}=\dfrac{x}{5-x}+\dfrac{5}{x-5}=\dfrac{x}{5-x}-\dfrac{5}{5-x}=\dfrac{x-5}{5-x}=\dfrac{-\left(5-x\right)}{5-x}=-1\)
Vậy A=-1
