`P=(3x^2+3x)/((x+1)(2x-6))`
`a)ĐKXĐ:(x+1)(2x-6)\ne0`
`→x\ne-1;x\ne3`
`b)` `P=1`
`→(3x^2+3x)/((x+1)(2x-6))=1`
`→3x^2+3x=(x+1)(2x-6)`
`→3x^2+3x=2x^2-4x-6`
`→3x^2-2x^2+3x+4x=-6`
`→x^2+7x+6=0`
`→(x+1)(x+6)=0`
`→` \(\left[ \begin{array}{l}x=-1(l)\\x=-6\end{array} \right.\)
Vậy `x=-6` thì `P=1`