d: \(=\dfrac{2x^2+10x+x^2-5x-20x}{\left(x+5\right)\left(x-5\right)}=\dfrac{3x^2-15x}{\left(x+5\right)\left(x-5\right)}=\dfrac{3x}{x+5}\)
\(e,\left(x^3+x^2+x+6\right):\left(x+2\right)\\ =\left[\left(x^3+2x^2\right)-\left(x^2+2x\right)+\left(3x+6\right)\right]:\left(x+2\right)\\ =\left[x^2\left(x+2\right)-x\left(x+2\right)+3\left(x+2\right)\right]:\left(x+2\right)\\ =\left[\left(x^2-x+3\right)\left(x+2\right)\right]:\left(x+2\right)\\ =x^2-x+3\)
\(f,\dfrac{2x}{x-5}+\dfrac{x}{x+5}-\dfrac{20x}{x^2-25}\\ =\dfrac{2x\left(x+5\right)+x\left(x-5\right)-20x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{2x^2+10x+x^2-5x-20x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{3x^2-15x}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{3x\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\\ =\dfrac{3x}{x+5}\)
