\(A=\left|2x-1\right|+5\ge5\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

\(B=\left|2x-2014\right|+2015\ge2015\forall x\)

Dấu '=' xảy ra khi x=1007

*\(x\ge\dfrac{1}{2}\Leftrightarrow\left|2x-1\right|=2x-1\)

\(D=\left(2x-1\right)^2-3\left(2x-1\right)+2=\left(2x-1\right)^2-2.\dfrac{3}{2}\left(2x-1\right)+\dfrac{9}{4}-\dfrac{1}{4}=\left(2x-1-\dfrac{3}{2}\right)^2-\dfrac{1}{4}=\left(2x-\dfrac{5}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)\(D_{min}=-\dfrac{1}{4}\Leftrightarrow x=\dfrac{5}{4}\left(1\right)\)

*\(x< \dfrac{1}{2}\Leftrightarrow\left|2x-1\right|=-2x+1\)

\(D=\left(2x-1\right)^2+3\left(2x-1\right)+2=\left(2x-1\right)^2+2.\dfrac{3}{2}\left(2x-1\right)+\dfrac{9}{4}-\dfrac{1}{4}=\left(2x-1+\dfrac{3}{2}\right)^2-\dfrac{1}{4}=\left(2x+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)\(D_{min}=-\dfrac{1}{4}\Leftrightarrow x=\dfrac{-1}{4}\left(2\right)\)
-Từ (1) và (2) suy ra \(D_{min}=-\dfrac{1}{4}\Leftrightarrow x\in\left\{\dfrac{5}{4};\dfrac{-1}{4}\right\}\)

a) | 2x - 1 | = 1- 3x

\(\orbr{\begin{cases}2x-1=1-3x\\2x-1=-\left(1-3x\right)\end{cases}}\)

\(\orbr{\begin{cases}2x-3x=1+1\\2x-1=-1+3x\end{cases}}\)

\(\orbr{\begin{cases}-x=2\\2x+3x=-1+1\end{cases}}\)

\(\orbr{\begin{cases}x=-2\\5x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=0\end{cases}}\)

b) | 1 - 2x | = x + 1 

\(\orbr{\begin{cases}1-2x=x+1\\1-2x=-\left(x+1\right)\end{cases}}\)

\(\orbr{\begin{cases}-2x-x=1-1\\-2x+x=-1-1\end{cases}}\)

\(\orbr{\begin{cases}-3x=0\\-x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

tương tự

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

\(=>2x+\frac{1}{2}=\left|x+1\right|\)

Mà \(\left|x+1\right|\ge0\) nên \(2x+\frac{1}{2}\ge0\)

\(=>2x\ge-\frac{1}{2}=>x\ge-\frac{1}{4}\)

\(=>2x+\frac{1}{2}=x+1=>x=\frac{1}{2}\)

\(x\ge-\frac{1}{2}\Rightarrow3x-2x-1=0\Rightarrow x=1\)

\(x< \frac{-1}{2}\Rightarrow3x+2x+1\Rightarrow x=-\frac{1}{5}\left(loai\right)\)

\(3x-|2x-1|=2\Leftrightarrow|2x-1|=2-3x\)

\(\Rightarrow-2x+1=2-3x\)hoặc \(-2x+1=3x-2\)

\(\Rightarrow1x+1=2\)hoặc \(-5x+1=-2\)

\(\Rightarrow x=1\)hoặc\(x=\frac{5}{3}\)

\(\left(2x-3\right)\left(6-2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\6-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}}\)

Vậy ...

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

\(\Leftrightarrow2.\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{4}\)

\(\Leftrightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=\frac{-7}{8}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{29}{12}\\x=\frac{-13}{12}\end{cases}}\)

Vậy..