\(A=\left|x-\dfrac{1}{2}\right|+\dfrac{3}{4}-x\)
Với \(x\ge\dfrac{1}{2}\) thì \(x-\dfrac{1}{2}\ge0\) nên \(\left|x-\dfrac{1}{2}\right|=x-\dfrac{1}{2},\) thay vào A ta có:
\(A=x-\dfrac{1}{2}+\dfrac{3}{4}-x=\dfrac{1}{4}\)
Với \(x< \dfrac{1}{2}\) thì \(x-\dfrac{1}{2}< 0\) nen \(\left|x-\dfrac{1}{2}\right|=-x+\dfrac{1}{2},\) thay vào A ta có:
\(A=-x+\dfrac{1}{2}+\dfrac{3}{4}-x=-2x+\dfrac{5}{4}\)
b) Với \(x\ge\dfrac{1}{2}\) thì \(A=\dfrac{1}{4}\) _______( 1 )_______
Với \(x< \dfrac{1}{2}\) thì \(-2x>-1\Leftrightarrow-2x+\dfrac{5}{4}>-1+\dfrac{5}{4}\) hay \(A>\dfrac{1}{4}\) __________( 2 )_________
Từ ( 1 ) và ( 2 ) suy ra \(A\ge\dfrac{1}{4}\)
Vậy \(GTNN\left(A\right)=\dfrac{1}{4}\) khi \(x\ge\dfrac{1}{2}\)