\(\left|x-1,5\right|+\left|3x-2,5\right|=0\)
Do \(\left|x-1,5\right|\ge0;\left|3x-2,5\right|\ge0\Rightarrow\left|x-1,5\right|+\left|3x-2,5\right|=0\Leftrightarrow\hept{\begin{cases}x=1,5\\3x=2,5\end{cases}}\Leftrightarrow x=\hept{\begin{cases}1,5\\\frac{5}{\frac{2}{3}}\end{cases}}\)(KTM)
Vậy...
\(\left|x-1,5\right|+\left|3x-2,5\right|=0\)
\(\Rightarrow x-1,5=0\) và \(3x-2,5=0\)
\(x=0+1,5\) \(3x=0+2,5\)
\(x=1,5\) \(3x=2,5\)
\(x=2,5:3\)
\(x=\frac{25}{30}\)
\(\Rightarrow x=1,5\) và \(x=\frac{25}{30}\) (vô lí)
\(\Rightarrow x\in\varnothing\)
