Ta có: |2x - 1| \(\ge\)0 \(\forall\)x
=> |2x - 1| + 5 \(\ge\)5 \(\forall\)x
Dấu "=" xảy ra <=> 2x - 1 = 0 <=> x = 1/2
Vậy MinA = 5 <=> x = 1/2
B = |0,5x + 1/3| - 7
Ta có: |0,5x + 1/3| \(\ge\)0 \(\forall\)x
=> |0,5x + 1/3| - 7 \(\ge\)-7 \(\forall\)x
Dấu "=" xảy ra <=> 0,5x + 1/3 = 0 <=> x = -2/3
Vậy MinB = -7 <=> x = -2/3
C = |x + 2| + |x + 3|
Ta có: C = |x + 2| + |x + 3|
C = |x + 2| + |-x - 3| \(\ge\)|x + 2 - x - 3| = |-1| = 1
Dấu "=" xảy ra <=> (x + 2)(-x - 3) \(\ge\)0
=>-3 \(\le\)x \(\le\)-2
Vậy MinC = 1 <=> -3 \(\le\)x \(\le\)-2