a) \(A=1,7+\left|3,4-x\right|\ge1,7\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left|3,4-x\right|=0\Rightarrow x=3,4\)
Vậy Min(A) = 1,7 khi x = 3,4
b) \(B=\left|x+2,8\right|-3,5\ge-3,5\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left|x+2,8\right|=0\Rightarrow x=-2,8\)
Vậy Min(B) = -3,5 khi x = -2,8
c) \(C=3,7+\left|4,3-x\right|\ge3,7\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left|4,3-x\right|=0\Rightarrow x=4,3\)
Vậy Min(C) = 3,7 khi x = 4,3
các câu khác thì sao?
d) \(C=\left|x+2\right|+\left|x-3\right|=\left|x+2\right|+\left|3-x\right|\)
\(\ge\left|x+2+3-x\right|=\left|5\right|=5\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x+2\right)\left(3-x\right)\ge0\Rightarrow-2\le x\le3\)
Vậy Min(C) = 5 khi \(-2\le x\le3\)
e) \(E=\left|2x-4\right|+\left|2x+5\right|=\left|4-2x\right|+\left|2x+5\right|\)
\(\ge\left|4-2x+2x+5\right|=\left|9\right|=9\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(4-2x\right)\left(2x+5\right)\ge0\Rightarrow-\frac{5}{2}\le x\le2\)
Vậy Min(E) = 9 khi \(-\frac{5}{2}\le x\le2\)
g) \(G=3\left|x-2\right|+\left|3x+1\right|=\left|6-3x\right|+\left|3x+1\right|\)
\(\ge\left|6-3x+3x+1\right|=\left|7\right|=7\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(6-3x\right)\left(3x+1\right)\ge0\Rightarrow-\frac{1}{3}\le x\le2\)
Vậy Min(G) = 7 khi \(-\frac{1}{3}\le x\le2\)
từ đâu mà có \(|x-3|=|3-x|\) vậy