\(\begin{array}{l}P = \left(\dfrac{3}{2x+4} + \dfrac{x}{2-x} + \dfrac{2x^2 + 3}{x^2 - 4}\right):\dfrac{2x-1}{4x-8}\quad \left(x\ne \pm 2;x \ne \dfrac12\right)\\a)\quad P = \dfrac{3(x-2)- 2x(x+2) + 2(2x^2 + 3)}{2(x-2)(x+2)}\cdot \dfrac{4(x-2)}{2x-1}\\\Leftrightarrow P = \dfrac{2(2x^2 -x)}{(x+2)(2x-1)}\\\Leftrightarrow P = \dfrac{2x(2x - 1)}{(x+2)(2x-1)}\\\Leftrightarrow P = \dfrac{2x}{x+2}\\b)\quad 4x^2 - 1 =0\\\Leftrightarrow x^2 = \dfrac14\\\Leftrightarrow x = \pm \dfrac12\\+)\quad x = - \dfrac12\ \text{thay vào P ta được:}\\\quad P = \dfrac{2\cdot\dfrac{1}{2}}{\dfrac{1}{2} + 2} = \dfrac{2}{5}\\+)\quad x = \dfrac12\ \text{(không thỏa ĐKXĐ)}\\\text{Do đó không tồn tại P khi $x = \dfrac12$}\\c)\quad P < 2\\\Leftrightarrow \dfrac{2x}{x+2} < 2\\\Leftrightarrow \dfrac{2x}{x+2} - 2 <0\\\Leftrightarrow \dfrac{2x-2(x+2)}{x+2}<0\\\Leftrightarrow \dfrac{-4}{x+2} <0\\\Leftrightarrow x+2 >0\\\Leftrightarrow x>-2\\\text{Kết hợp điều kiện xác định ta được:}\\x > -2;\ x \ne 2;\ x \ne \dfrac12\end{array}\)

a/ ĐKXĐ: $x\ne \pm 2,x\ne \dfrac{1}{2}$

$P\,=\left(\dfrac{3}{2x+4}+\dfrac{x}{2-x}+\dfrac{2x^2+3}{x^2-4}\right):\dfrac{2x-1}{4x-8}\\\quad=\left(\dfrac{3(x-2)}{2(x+2)(x-2)}-\dfrac{2(x+2)x}{2(x+2)(x-2)}+\dfrac{2(2x^2+3)}{2(x-2)(x+2)}\right).\dfrac{4(x-2)}{2x-1}\\\quad =\dfrac{3x-6-2x^2-4x+4x^2+6}{2(x-2)(x+2)}.\dfrac{4(x-2)}{2x-1}\\\quad =\dfrac{2x^2-x}{x+2}.\dfrac{2}{2x-1}\\\quad =\dfrac{x(2x-1)}{x+2}.\dfrac{2}{2x-1}\\\quad =\dfrac{2x}{x+2}$

$→$ ĐPCM

Vậy ta có điều phải chứng minh

b/ $4x^2-1=0\\↔(2x-1)(2x+1)=0\\↔\left[\begin{array}{1}2x-1=0\\2x+1=0\end{array}\right.\\↔\left[\begin{array}{1}2x=1\\2x=-1\end{array}\right.\\↔\left[\begin{array}{1}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{array}\right.$

mà theo ĐKXĐ: $x\ne \pm 2,x\ne \dfrac{1}{2}$

$→x=-\dfrac{1}{2}$

Với $x=-\dfrac{1}{2}$ (thỏa mãn điều kiện)

$→P=\dfrac{2.\left(-\dfrac{1}{2}\right) }{-\dfrac{1}{2}+2}=\dfrac{-1}{\dfrac{3}{2}}=-\dfrac{2}{3}$

Vậy $P=-\dfrac{2}{3}$ với $4x^2-1=0$

c/ $P<2\\↔\dfrac{2x}{x+2}<2\\↔\dfrac{2x}{x+2}-2<0\\↔\dfrac{2x-2(x+2)}{x+2}<0\\↔\dfrac{2x-2x-4}{x+2}<0\\↔\dfrac{-4}{x+2}<0\\→x+2>0\\↔x>-2\left(x\ne 2;\dfrac{1}{2}\right)$

Vậy $x>-2\left(x\ne 2;\dfrac{1}{2}\right)$ thì $P<2$