$(x^2+x-6)(x^2+x+2)$ $(1)$
Đặt: $x^2+x-6=t$
$\to$ $\begin{cases}x^2+x-6=t-4\\x^2+x+2=t+4\end{cases}$
$pt(1):(t-4)(t+4)$
$\to t^2-16$
Hay: $(x^2+x-2)^2-16$
Vì $(x^2+x-2)^2 \geq 0∀x$
$\to (x^2+x-12)^2-16≥-16 ∀x$
Dấu "=" xảy ra $⇔(x^2+x-2)^2=0⇔x^2+x-2=0⇔$ \(\left[ \begin{array}{l}x=1\\x=2\end{array} \right.\)
Vậy $pt (1)$ đạt gtnn là $-16⇔$\(\left[ \begin{array}{l}x=1\\x=2\end{array} \right.\)
