Bài 3:
a: Ta có: \(P=\dfrac{x}{x+2}+\dfrac{2}{x-2}+\dfrac{2x+4}{4-x^2}\)
\(=\dfrac{x^2-2x+2x+4-2x-4}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{x}{x+2}\)
b: Ta có: |x+1|=3
\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(loại\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
Thay x=-4 vào P, ta được:
\(P=\dfrac{-4}{-4+2}=\dfrac{-4}{-2}=2\)