Ta có:
\(A=x^4+2x^3+9x^2+8x+27\)
\(\Leftrightarrow A=x^4+x^2+16+2x^3+8x+8x^2+11\)
\(\Leftrightarrow A=\left(x^2+x+4\right)^2+11\)
\(\Leftrightarrow A=\left(x^2+x+\dfrac{1}{4}+\dfrac{15}{4}\right)^2+11\)
\(\Leftrightarrow A=\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}\right]^2+11\)
\(\ge\left(\dfrac{15}{4}\right)^2+11=\dfrac{401}{16}\)
Vậy \(A_{min}=\dfrac{401}{16}\), đạt được khi \(x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)
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