\(A=3\left|2x+1\right|-\frac{5}{3}\)
Vì \(\left|2x+1\right|\ge0,\forall x\)
\(\Rightarrow3\left|2x+1\right|\ge0,\forall x\)
\(\Rightarrow3\left|2x+1\right|-\frac{5}{3}\ge\frac{-5}{3},\forall x\)
\(\Rightarrow A\ge\frac{-5}{3}\)
Dấu " = " xảy ra khi:
\(\left|2x+1\right|=0\)
\(2x+1=0\)
\(2x=0-1\)
\(2x=-1\)
\(x=-1:2\)
\(x=\frac{-1}{2}\)
Vậy \(A_{min}=\frac{-5}{3}khi\)\(x=\frac{-1}{2}\)
\(B=-\left|3,5+2x\right|+5\)
Vì \(\left|3,5+2x\right|\ge0,\forall x\)
\(\Rightarrow-\left|3,5+2x\right|\le0,\forall x\)
\(\Rightarrow-\left|3,5+2x\right|+5\le5\)
\(\Rightarrow B\le5\)
Dấu " = " xảy ra khi:
\(3,5+2x=0\)
\(2x=3,5-0\)
\(2x=3,5\)
\(x=3,5:2\)
\(x=\frac{3,5}{2}\)
Vậy \(B_{max}=5khi\)\(x=\frac{3,5}{2}\)
