`f(x)=x^2+(m+1)x+2m+3`

`f(x)>0 AA x` khi:

`{(a>0),(Delta<0):}`

`-> {(1>0),((m+1)^2-4*1(2m+3)<0):}`

`-> m^2+2m+1-8m-12<0`

`-> m^2-6m-11<0`

`g(x)=m^2-6m-11` có hai nghiệm `m=3-2sqrt5`, `m=3+2sqrt5`

Bảng xét dấu:

\begin{array}{|c|cc|}\hline m&-\infty&&3-2\sqrt{5}&&3+2\sqrt{5}&&+\infty\\\hline f(m)&&+&0&-&0&+&\\\hline\end{array}

`-> g(x)<0 -> m in (3-2sqrt5; 3+2sqrt5)`

`-> f(x)>0 AA x -> m in (3-2sqrt5;3+2sqrt5)`