Ta có : \(\left|4,3-x\right|\ge0=>3,7+\left|4,3-x\right|\ge3,7\)
Dấu "=" xảy ra khi \(4,3-x=0=>x=4,3\)
Vậy \(A_{min}=3,7\)khi \(x=4,3\)
Vì \(|4,3-x|\ge0;\forall x\)
\(\Rightarrow3,7+|4,3-x|\ge3,7+0;\forall x\)
Hay \(A\ge3,7;\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow|4,3-x|=0\)
\(\Leftrightarrow x=4,3\)
Vậy MIN A =3,7 \(\Leftrightarrow x=4,3\)
A = 3,7 + |4,3 - x| \(\ge\)3,7
Amin = 3,7 khi 4,3 - x = 0 <=> x = 4,3
#hatsunemiku
Vì \(\left|4,3-x\right|\ge0\Rightarrow3,7+\left|4,3-x\right|\ge3,7\)
Dấu " = " xảy ra khi :
\(\left|4,3-x\right|=0\)
\(\Rightarrow4,3-x=0\)
\(\Rightarrow x=4,3\)
Vậy \(A_{min}=3,7\)khi \(x=4,3\)