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[tex]\dfrac{-2x+5}{x+4} - \dfrac{8}{x+3}\\ \\= \dfrac{(-2x+5)(x+3)-8(x+4)}{(x+4)(x+3)}\\\\=\dfrac{-2x^2-6x+5x+15 -8x -32}{x^2+3x+4x+12}\\\\=\dfrac{-2x^2-9x-17}{x^2+7x+12}\\\\\\=\dfrac{-2x^2+(-9)x +(-17)}{x^2+7x+12}\\\\\text{Hence, }~ A=-2,~ B=-9, ~C=-17, ~D=12[/tex]