\(3|x-\frac{1}{2}|+\frac{3}{4}=-2|x-\frac{1}{2}|\)
\(\Rightarrow\) \(3|x-\frac{1}{2}|+2|x-\frac{1}{2}|=-\frac{3}{4}\)
\(\Rightarrow5|x-\frac{1}{2}|=-\frac{3}{4}\)
\(\Rightarrow|x-\frac{1}{2}|=-\frac{3}{4}:5=-\frac{3}{20}\) ( vô lý )
Vậy ko tồn tại x thỏa mãn yêu cầu bài toán
\(3.\left|x-\frac{1}{2}\right|+\frac{3}{4}=-2.\left|\frac{1}{2}-x\right|\)
\(x-\frac{1}{2}\ge0\) cho: \(x\ge\frac{1}{2}\) do đó: \(x\ge\frac{1}{2};\left|x-\frac{1}{2}\right|=x-\frac{1}{2}\)
\(x-\frac{1}{2}< 0\) cho: \(x< \frac{1}{2}\) do đó: \(x\le\frac{1}{2};\left|x-\frac{1}{2}\right|=-\left(x-\frac{1}{2}\right)\)
\(\frac{1}{2}-x\ge0\) cho \(x\le\frac{1}{2}\) do đó: \(x\le\frac{1}{2};\left|\frac{1}{2}-x\right|=\frac{1}{2}-x\)
\(\frac{1}{2}-x< 0\) cho \(x>\frac{1}{2}\) do đó: \(x>\frac{1}{2}\left|\frac{1}{2}-x\right|=-\left(\frac{1}{2}-x\right)\)
\(x< \frac{1}{2};x\ge\frac{1}{2}\)
Ta xét 2th:
Th1: \(3\left[-\left(x-\frac{1}{2}\right)\right]+\frac{3}{4}=-2\left(\frac{1}{2}-x\right)\)
\(x=\frac{13}{20}\) (loại)
Th2: \(3\left(x-\frac{1}{2}\right)+\frac{3}{4}=2\left[-\left(\frac{1}{2}-x\right)\right]\)
\(x=\frac{7}{20}\) (loại)
=> Không có giá trị thỏa mãn đề bài.