Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(D=\left|2x+2,5\right|+\left|2x-3\right|\)
\(=\left|2x+2,5\right|+\left|3-2x\right|\)
\(\ge\left|2x+2,5+3-2x\right|=\frac{11}{2}\)
Dấu "=" xảy ra khi \(-1,25\le x\le1,5\)
Mà x nguyên suy ra \(x\in\left\{-1;0;1\right\}\)
Vậy \(x\in\left\{-1;0;1\right\}\) thì D đạt giá trị nhỏ nhất
Ta có: \(D=\left|2x+2,5\right|+\left|2x-3\right|=\left|2x+2,5\right|+\left|3-2x\right|\)
Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(A=\left|2x+2,5\right|+\left|3-2x\right|\ge\left|2x+2,5+3-2x\right|=5,5\)
Dấu " = " xảy ra khi \(2x+2,5\ge0;3-2x\ge0\)
\(\Rightarrow x\ge-1,25;x\le1,5\)
\(\Rightarrow-1,25\le x\le1,5\)
Mà \(x\in Z\)
\(\Rightarrow x\in\left\{-1;0;1\right\}\)
Vậy \(MIN_D=5,5\) khi \(x\in\left\{-1;0;1\right\}\)