$\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}$ $=\frac{\left(x-3\right)^2}{\left(x+3\right)^2\left(x-3\right)^2}+\frac{-x^2-6x-9}{\left(x+3\right)^2\left(x-3\right)^2}+\frac{x\left(x+3\right)\left(x-3\right)}{\left(x+3\right)^2\left(x-3\right)^2}$ $=\frac{\left(x-3\right)^2-x^2-6x-9+x\left(x+3\right)\left(x-3\right)}{\left(x+3\right)^2\left(x-3\right)^2}$ $=\frac{x^3+3^2-12x-3^2x-9}{\left(x+3\right)^2\left(x-3\right)^2}$ $=\frac{x^3-21x}{\left(x+3\right)^2\left(x-3\right)^2}$
Mong cho lời giải hay nhất ❤️⭐️
