If $\frac{x^2+5x+7}{(x-3)^3}=\frac{A}{(x-3)}+\frac{B}{(x-3)^2}+\frac{C}{(x-3)^3}$,then $9A-3B+C=$

  • A
    $2$
  • B
    $5$
  • C
    $7$
  • D
    $9$

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