\(a,\dfrac{9-\left(x+5\right)^2}{x^2+4x+4}=\dfrac{\left(3-x-5\right)\left(3+x+5\right)}{\left(x+2\right)^2}=\dfrac{\left(-x-2\right)\left(x+8\right)}{\left(x+2\right)^2}=\dfrac{-\left(x+2\right)\left(x+8\right)}{\left(x+2\right)^2}=\dfrac{-\left(x+8\right)}{x+2}\\ b,\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{16x}=\dfrac{\left(x+2-x+2\right)\left(x+2+x-2\right)}{16x}=\dfrac{4.2x}{16x}=\dfrac{8x}{16x}=\dfrac{1}{2}\)
\(c,\dfrac{5x^3+5x}{x^4-1}=\dfrac{5x\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2-1\right)}=\dfrac{5x}{x^2-1}=\dfrac{5x}{\left(x-1\right)\left(x+1\right)}\\ d,\dfrac{x^7-x^4}{x^6-1}=\dfrac{x^4\left(x^3-1\right)}{\left(x^3-1\right)\left(x^3+1\right)}=\dfrac{x^4}{x^3+1}\)
