\(P=\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^2-3x}{2x^2-x^3}\\ P=\left(\dfrac{\left(2+x\right)^2}{\left(2-x\right)\left(2+x\right)}-\dfrac{4x^2}{\left(2-x\right)\left(2+x\right)}-\dfrac{\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\\ P=\dfrac{4+4x+x^2-4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}.\dfrac{x\left(2-x\right)}{\left(x-3\right)}\\ P=\dfrac{-3x^2+8x}{\left(2-x\right)\left(2+x\right)}.\dfrac{x\left(2-x\right)}{\left(x-3\right)}\\ P=\dfrac{x\left(-3x^2+8x\right)}{\left(2+x\right)\left(x-3\right)}\)
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