\(=\dfrac{1}{x-3}-\dfrac{3}{2\left(x+3\right)}-\dfrac{x}{2\left(x-3\right)^2}\left(x\ne\pm3\right)\\ =\dfrac{2\left(x-3\right)\left(x+3\right)-3\left(x-3\right)\left(x+3\right)-x\left(x+3\right)}{2\left(x-3\right)^2\left(x+3\right)}\\ =\dfrac{-x^2+9-x^2-3x}{2\left(x-3\right)^2\left(x+3\right)}=\dfrac{-2x^2-3x+9}{2\left(x-3\right)^2\left(x+3\right)}\\ =\dfrac{\left(x+3\right)\left(3-2x\right)}{2\left(x-3\right)^2\left(x+3\right)}=\dfrac{3-2x}{2\left(x-3\right)^2}\)
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