Question

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Answer 1

\(\dfrac{1}{-3x^2+18x-27}=\dfrac{-1}{3\left(x-3\right)^2}=\dfrac{-\left(x+1\right)}{3\left(x-3\right)^2\left(x+1\right)}\\ \dfrac{1}{3x^2-6x-9}=\dfrac{1}{3\left(x-3\right)\left(x+1\right)}=\dfrac{x-3}{3\left(x-3\right)^2\left(x+1\right)}\)