a) ĐK : x + 1 $\neq$ 0 <=> x $\neq$ -1
2x - 6 $\neq$ 0 <=> x $\neq$ 3
b) Ta có: P = 1
<=> 3x²+ 3x = (x+1)(2x - 6)
<=> 3x² + 3x = 2x² - 6x + 2x - 6
<=> 3x² + 3x - 2x² + 4x + 6 = 0
<=> x² + 7x + 6 = 0
<=> x² + x + 6x + 6 = 0
<=> (x+1)(x+6) = 0
TH1: x + 1 = 0 <=> x = -1 (KTM)
TH2: x + 6 = 0 <=> x = -6