\(B=\left(x^2+x\right)^2+4\left(x^2+x\right)+4-16=\left(x^2+x+2\right)^2-16\ge-16\)
Dấu \("="\Leftrightarrow x^2+x+2=0\Leftrightarrow x\in\varnothing\left(x^2+x+2>0\right)\)
Vậy dấu \("="\) ko xảy ra nên sẽ ko tính đc GTNN
\(B=\left(x^2+x\right)^2+4\left(x^2+x^2\right)-12\)
\(=\left(x^2+x\right)^2+4\left(x^2+x\right)+4-16\)
\(=\left(x^2+x+2\right)^2-16\)
\(=\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right]^2-16\)
Do \(\left(x+\dfrac{1}{2}\right)^2\ge0;\forall x\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
\(\Rightarrow B\ge\left(\dfrac{7}{4}\right)^2-16=-\dfrac{207}{16}\)
\(B_{min}=-\dfrac{207}{16}\) khi \(x=-\dfrac{1}{2}\)
