\(B=\left(1-2x\right)\left(x+3\right)-9\)
\(B=x+3-2x^2-6x-9\)
\(B=-2x^2-5x-6\)
\(B=-2\left(x^2+\dfrac{5}{2}x+3\right)\)
\(B=-2\left(x^2+2x\dfrac{5}{4}+\dfrac{25}{16}-\dfrac{25}{16}+3\right)\)
\(B=-2\left[\left(x+\dfrac{5}{4}\right)^2+\dfrac{23}{16}\right]\)
\(B=-2\left(x+\dfrac{5}{4}\right)^2-\dfrac{23}{8}\le-\dfrac{23}{8}\forall x\in R\)
dấu = xảy ra khi \(x+\dfrac{5}{4}=0\Leftrightarrow x=-\dfrac{5}{4}\)
vậy \(B_{max}=-\dfrac{23}{8}\) khi \(x=-\dfrac{5}{4}\)
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