\(\begin{array}{l}
P = \left(\dfrac{x^2}{x^2 - 4} - \dfrac{x}{x-2} +\dfrac{2}{x+2}\right)\cdot\dfrac{x+2}{2}\\
a)\\
\text{Điều kiện xác định:}\\
\begin{cases}x^2 - 4 \ne 0\\x - 2\ne 0\\x + 2\ne 0\end{cases}\Leftrightarrow \begin{cases}x\ne -2\\x \ne 2\end{cases}\\
b)\\
P = \left[\dfrac{x^2}{(x-2)(x+2)} - \dfrac{x(x+2)}{(x-2)(x+2)} + \dfrac{2(x-2)}{(x-2)(x+2)}\right]\cdot \dfrac{x+2}{2}\\
\to P = \dfrac{x^2 - x^2 - 2x + 2x - 4}{(x-2)(x+2)}\cdot \dfrac{x+2}{2}\\
\to P = \dfrac{-4}{(x-2)(x+2)}\cdot \dfrac{x+2}{2}\\
\to P = \dfrac{-2}{x-2}\\
c)\\
\text{Thay $x=20$ vào $P$ ta được:}\\
P = \dfrac{-2}{20 - 2} = \dfrac{-2}{18} = - \dfrac19\\
d)\\
\quad P = 4\\
\to \dfrac{-2}{x - 2} = 4\\
\to 4(x-2) = -2\\
\to 4x - 8 = -2\\
\to 4x = 6\\
\to x = \dfrac32
\end{array}\)