Giải:
\(A=\left|2,5-x\right|+\left(4,7+1,1\right)\)
\(\Leftrightarrow A=\left|2,5-x\right|+5,8\)
Vì \(\left|2,5-x\right|\ge0;\forall x\)
\(\Leftrightarrow\left|2,5-x\right|+5,8\ge5,8;\forall x\)
\(\Leftrightarrow A\ge5,8;\forall x\)
\(\Leftrightarrow A_{Min}=5,8\)
Dấu "=" xảy ra:
\(\Leftrightarrow2,5-x=0\Leftrightarrow x=2,5\)
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