F = \(\left[\left(x+2\right)\left(x+3\right)-1\right]\) . \(\left[\left(x+2\right)\left(x+3\right)+1\right]\)
= \(\left(x+2\right)^2\left(x+3\right)^2-1\) \(\ge-1\)
Dấu "=" xảy ra <=> \(\left(x+2\right)^2\left(x+3\right)^2\) =0
<=> (x+2)(x+3) = 0
<=> \(\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
Chắc zậy