\( \begin{array}{l}
a)\\ \\ĐKXĐ:x\ne0;x\ne±2\\ \\
P=(\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}-\dfrac{4x^3}{x^3-4x}):\dfrac{x-3}{2x-x^2}\\ \\
=(\dfrac{x(x+2)^2}{x(x+2)(2-x)}-\dfrac{x(2-x)^2}{x(x+2)(2-x)}+\dfrac{4x^3}{x(x+2)(2-x)}).\dfrac{2x-x^2}{x-3}\\ \ \\
=(\dfrac{x^3+4x^2+4x-4x+4x^2-x^3+4x^3}{x(x+2)(2-x)}).\dfrac{2x-x^2}{x-3}\\ \\
=\dfrac{4x^3+8x^2}{x(x+2)(2-x)}.\dfrac{x(2-x)}{x-3}\\ \\
=\dfrac{4x^2(x+2)x(2-x)}{x(x+2)(2-x)(x-3)}\\ \\
=\dfrac{4x^2}{x-3}\\ \\
b)\\ \\
\text{Thay x=1 vào P, ta được:}\\ \\
P=\dfrac{4.1^2}{1-3}=-2\\ \\
\text{Thay x=-1 vào P, ta được:}\\ \\
P=\dfrac{4.(-1)^2}{-1-3}=-1\\ \\
c)\\ \\
P>0\\ \\
⇔x-3>0(do\ 4x^2>0∀x\ne \left\{ {0; ± 2} \right\}\\ \\
⇔x>3\\ \\
d)\\ \\
\dfrac{1}{P}=\dfrac{x-3}{4x^2}\\
⇒\dfrac{1}{P}-1=\dfrac{x-3-4x^2}{4x^2}=\dfrac{-4(x-\dfrac{1}{8})^2-\dfrac{47}{16}}{4x^2}<0∀x\ne0\\ \\
⇒\dfrac{1}{P}<1
\end{array} \)
