Giải:
\(A=\left(2x+\dfrac{1}{3}\right)^4-1\)
\(\Leftrightarrow A=\left(2x+\dfrac{1}{3}\right)^4-1\ge-1\)
\(\Leftrightarrow A_{Min}=-1\)
\(\Leftrightarrow2x+\dfrac{1}{3}=0\)
\(\Leftrightarrow2x=-\dfrac{1}{3}\)
\(\Leftrightarrow x=-\dfrac{1}{6}\)
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