a)Đk:\(4x^2-1\ne0\Rightarrow\left(2x-1\right)\left(2x+1\right)\ne0\)\(\Rightarrow\begin{cases}x\ne\frac{1}{2}\\x\ne-\frac{1}{2}\end{cases}\)
b)Với \(P=0\Rightarrow\frac{\left(2x+1\right)x}{4x^2-1}=0\Rightarrow\frac{\left(2x+1\right)x}{\left(2x+1\right)\left(2x-1\right)}=0\)
\(\Rightarrow\frac{x}{2x-1}=0\Rightarrow x=0\) (thỏa mãn)
Vậy với x=0 thì P=0