Ta có: \(A=2,5+\left|x-3\right|\ge2,5\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left|x-3\right|=0\)
\(\Leftrightarrow x-3=0\Rightarrow x=3\)
Vậy Min(A) = 2,5 khi x = 3
A = 2,5 + | x - 3 |
| x - 3 | ≥ 0 ∀ x => 2, 5 + | x - 3 | ≥ 2, 5
Dấu "=" xảy ra khi x = 3
=> MinA = 2,5 <=> x = 3
B = -2, 5 - | 3x - 1 |
-| 3x - 1 | ≤ 0 ∀ x => -2,5 - | 3x - 1 | ≤ -2, 5
Dấu "=" xảy ra khi x = 1/3
=> MaxB = -2, 5 <=> x = 1/3
C = -| x - 4 | + 2
-| x - 4 | ≤ 0 ∀ x => -| x - 4 | + 2 ≤ 2
Dấu "=" xảy ra khi x = 4
=> MaxC = 2 <=> x = 4
D = | 4, 2 - x | + 1
| 4, 2 - x | ≥ 0 ∀ x => | 4, 2 - x | + 1 ≥ 1
Dấu "=" xảy ra khi x = 4, 2
=> MinD = 1 <=> x = 4, 2
Ta có: \(B=-2,5-\left|3x-1\right|\le-2,5\left(\forall x\right)\)
Dấu "='' xảy ra khi: \(\left|3x-1\right|=0\)
\(\Leftrightarrow3x-1=0\Rightarrow x=\frac{1}{3}\)
Vậy Max(B) = -2,5 khi x = 1/3
Ta có: \(C=-\left|x-4\right|+2\le2\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left|x-4\right|=0\Rightarrow x=4\)
Vậy Max(C) = 2 khi x = 4
Ta có: \(D=\left|4,2-x\right|+1\ge1\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left|4,2-x\right|=0\Rightarrow x=4,2\)
Vậy Min(D) = 1 khi x = 4,2
A = 2,5 + | x - 3 |
+) Nếu x-3 > 0 => A> 2,5
+) Nếu x-3 < 0 => A< 2,5
Min A <=> x - 3> 0 ; x-3 đạt Min
=> x-3=1
=> x=4
Vậy A Min = 3,5 <=> x=4