\(A=x\left(x-1\right)\left(x+2\right)\left(x-3\right)-9\)
\(=\left(x^2-x\right)\left(x^2+2x-3x-6\right)-9\)
\(=\left(x^2-x\right)\left(x^2-x-6\right)-9\)
\(=\left(x^2-x-3+3\right)\left(x^2-x-3-3\right)-9\)
\(=\left(x^2-x-3\right)^2-9-9\)
\(=\left(x^2-x-3\right)^2-18\ge-18\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x^2-x-3=0\)
\(\Leftrightarrow x^2-x+\dfrac{1}{4}-\dfrac{13}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{13}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{\sqrt{13}}{2}\\x-\dfrac{1}{2}=\dfrac{-\sqrt{13}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{13}+1}{2}\\x=\dfrac{1-\sqrt{13}}{2}\end{matrix}\right.\)
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