Question

Tìm x biết:

\(^{\left|x-2\right|+\left|3-2x\right|=2x-1}\)

Answer 1

 I x - 2 I + I 3 - 2x I = 2x - 1 

I x - 2 + 3 - 2x I = 2x - 1

I -x + 1 I = 2x - 1

x + 1 = 2x - 1 

x + 1 + 1 = 2x

x + 2 = 2x 

x + 2 = x + x 

=> x = 2

Answer 2

Ta có: |x−2|+|3−2x|=2x−1 \(\Rightarrow\) |x−2|+|3−2x|\(\ge\) |x−2+3−2x| = |1-2x|

Answer 3

Ta có \(\left|x-2\right|=\hept{\begin{cases}x-2;x\ge2\\-x+2;x< 2\end{cases}}\)  ; \(\left|3-2x\right|=\hept{\begin{cases}3-2x;x\le\frac{3}{2}\\2x-3;x>\frac{3}{2}\end{cases}}\)

Vậy nên ta xét các trường hợp sau:

Với \(x\le\frac{3}{2}\), ta có \(-x+2+3-2x=2x-1\)

\(\Leftrightarrow5x=6\Leftrightarrow x=\frac{6}{5}\left(tm\right)\)

Với \(\frac{3}{2}< x< 2\), ta có: \(-x+2+2x-3=2x-1\)

\(\Leftrightarrow x=0\left(l\right)\)

Với \(x\ge2\), ta có \(x-2+2x-3=2x-1\)

\(\Leftrightarrow x=4\left(tm\right)\)

Vậy \(x\in\left\{\frac{6}{5};4\right\}\)