Giải thích các bước giải:
`(x^2+3x+2)(x^2+7x+12)=24`
`=>(x+1)(x+2)(x+3)(x+4)=24`
`=>[(x+1)(x+4)].[(x+2)(x+3)]=24`
`=>(x^2+5x+4)(x^2+5x+6)=24`
`=>(x^2+5x+5)^2-1=24`
`=>(x^2+5x+5)^2-25=0`
`=>(x^2+5x+5+5)(x^2+5x+5-5)=0`
`=>(x^2+5x)(x^2+5x+10)=0`
Mà `x^2+5x+10>0AAx`
`=>x^2+5x=0`
`=>x(x+5)=0`
`=>`\(\left[ \begin{array}{l}x=0\\x=-5\end{array} \right.\)
