B=Ix-2I+3 khong co gia tri lon nhat chi co GTNN=3

|3x-1|=7/6

=>3x-1=-7/6 hoặc 7/6

=>3x=-1/6

=>x=-1/18

=>3x=13/6

=>x=13/18

b)5/3*|x-1/2|+1/3=4/3

=>5/3*|x-1/2|=1

=>|x-1/2|=3/5

=>x-1/2=-3/5 hoặc 3/5

=>x=-1/10

=>x=11/10

-3|x - 1| = 9

   |x - 1| = 9 : (-3)

   |x - 1| = -3

=> x - 1 = -3

     x      = -3 + 1

     x      = -2

=> x - 1 = 3

     x      = 3 + 1

     x      = 4

Vậy x = -2 hoặc x = 4

- 3 . | x - 1 | = 9

=> | x - 1 | = 9 : ( - 3 ) = - 3

=> | x - 1 | = - 3 ( vô lí vì | x - 1 | ≥ 0 )

=> x ∈ ∅

|x-3| = |x-2|

TH1: x-3 = x-2

=>  x -x = -2 + 3

0 = 1 ( vô lí)

=> không tìm được x

TH2: x-3 = -x+2

=> x + x = 2 + 3

2x = 5

x = 5/2

KL:...

câu b lm tương tự 

\(\left|x-3\right|=\left|x-2\right|\)

TH1: \(x-3=x-2\Leftrightarrow0x=1\) (vô lí)

TH2: \(x-3=-\left(x-2\right)\Leftrightarrow x-3=-x+2\Leftrightarrow2x=5\Leftrightarrow x=2,5\)

Vậy x = 2,5

\(\left|5-x\right|=\left|7-x\right|\)

TH1: \(5-x=7-x\Leftrightarrow0x=2\)(vô lí)

TH2: \(5-x=-\left(7-x\right)\Leftrightarrow5-x=x-7\Leftrightarrow-2x=-12\Leftrightarrow x=6\)

Vậy x = 6

\(A=\left|x+\frac{3}{2}\right|\)

Vì \(\left|x+\frac{3}{2}\right|\ge0\)

Vậy \(GTNN_A=0\)tại \(x=\frac{-3}{2}\)

\(B=\left|x-\frac{1}{2}\right|+\frac{3}{4}\)

Vì \(\left|x-\frac{1}{2}\right|\ge0\)nên \(\left|x-\frac{1}{2}\right|+\frac{3}{4}\ge\frac{3}{4}\)

Vậy \(GTNN_B=\frac{3}{4}\)tại \(x=\frac{1}{2}\)

\(A=\left|x+\frac{3}{2}\right|=x+\frac{3}{2}\)

\(A=\left|x+\frac{3}{2}\right|\ge0\)

\(\Rightarrow GTNN\left(A\right)=0\)

\(B=\left|x-\frac{1}{2}\right|+\frac{3}{4}\ge\frac{3}{4}\)

\(\Rightarrow GTNN\left(B\right)=\frac{3}{4}\)