2x-1=4
2x-5=4
2x=4+1
2x=4+5
2x=5
2x=9
x=5/2
x=9/2
x=2.5
x=4.5
\(\left|2x-1\right|+\left|2x-5\right|=4\)
\(\Leftrightarrow\left|2x-1\right|+\left|5-2x\right|=4\)
Ta có: \(\hept{\begin{cases}\left|2x-1\right|\ge2x-1\forall x\\\left|5-2x\right|\ge5-2x\forall x\end{cases}}\)
\(\Rightarrow\left|2x-1\right|+\left|2x-5\right|\ge\left(2x-1\right)+\left(5-2x\right)=2x-1+5-2x=4\)
Mà \(\left|2x-1\right|+\left|2x-5\right|=4\)
\(\Rightarrow\hept{\begin{cases}\left|2x-1\right|=2x-1\\\left|5-2x\right|=5-2x\end{cases}\Leftrightarrow\hept{\begin{cases}2x-1\ge0\\5-2x\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge\frac{1}{2}\\x\le\frac{5}{2}\end{cases}\Rightarrow}\frac{1}{2}\le x\le\frac{5}{2}}\)
Vậy \(\frac{1}{2}\le x\le\frac{5}{2}\)
Tham khảo nhé~
