`|1/5x-0,6|=|1/2x+0,4|`
`@TH1:1/5x-0,6=1/2x+0,4`
`=>1/5x-1/2x=0,4+0,6`
`=>-3/10x=1`
`=>x=1:(-3/10)=-10/3`
`@TH2:1/5x-0,6=-1/2x-0,4`
`=>1/5x+1/2x=-0,4+0,6`
`=>7/10x=0,2=1/5`
`=>x=1/5:(7/10)=2/7`
Vậy `x in {-10/3;2/7}`
`|1/5x-0,6|=|1/2x+0,4|`
`|1/5x-3/5|=|1/2x+2/5|`
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{5}x-\dfrac{3}{5}=\dfrac{1}{2}x+\dfrac{2}{5}\\\dfrac{1}{5}x-\dfrac{3}{5}=\dfrac{2}{5}-\dfrac{1}{2}x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left(\dfrac{1}{5}+\dfrac{1}{2}\right)x=\dfrac{3}{5}-\dfrac{2}{5}\\\left(\dfrac{1}{5}-\dfrac{1}{2}\right)x=\dfrac{2}{5}+\dfrac{3}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{7}{10}x=\dfrac{1}{5}\\-\dfrac{3}{10}x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}:\dfrac{7}{10}\\x=1:\dfrac{\left(-3\right)}{10}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}.\dfrac{10}{7}\\x=1.\dfrac{\left(-10\right)}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{7}\\x=\dfrac{\left(-10\right)}{3}\end{matrix}\right.\)
Vậy....
`#BaoL i n h`
