Question

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Answer 1

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

\(\Rightarrow\dfrac{18}{\left(x-3\right)\left(x^2-9\right)}-\dfrac{3}{x^2-2x.3+3^2}-\dfrac{x}{x^2-9}=\dfrac{18}{\left(x^2-9\right)\left(x-3\right)}-\dfrac{3}{\left(x-3\right)^2}\)

\(\Rightarrow\dfrac{18}{\left(x-3\right)\left(x^2-9\right)}-\dfrac{3}{\left(x-3\right)^2}-\dfrac{x}{x^2-9}=\dfrac{18}{\left(x-3\right)\left(x^2-9\right)}-\dfrac{3}{\left(x-3\right)^2}\)

\(\Rightarrow-\dfrac{x}{x^2-9}=0\)

Xảy ra khi \(x=0\)

Vậy \(x=0.\)

Answer 2

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