$\begin{array}{l}a,\\\text{Điều kiện xác định:}\\\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}\leftrightarrow \begin{cases}x\ne-1\\x\ne3\end{cases}\\b,\\P=\dfrac{3x^2+3x}{(x+1)(2x-6)}\\=\dfrac{3x(x+1)}{(x+1)(2x-6)}\\=\dfrac{3x}{2x-6}\\P=1\leftrightarrow \dfrac{3x}{2x-6}=1\\\leftrightarrow 3x=2x-6\\\leftrightarrow x=-6\\\text{Vậy $x=-6$ thì $P=1$}\end{array}$
