\(\left|\dfrac{4}{5}x+\dfrac{1}{3}\right|=\left|\dfrac{2}{5}x+\dfrac{1}{3}\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{4}{5}x+\dfrac{1}{3}=\dfrac{2}{5}x+\dfrac{1}{3}\\\dfrac{4}{5}x+\dfrac{1}{3}=-\dfrac{2}{5}x-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{5}x=0\\\dfrac{6}{5}x=-\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{9}\end{matrix}\right.\)
`|4/5x+1/3|=|2/5x+1/3|`
\(=>\left[{}\begin{matrix}\dfrac{4}{5}.x+\dfrac{1}{3}=\dfrac{2}{5}x+\dfrac{1}{3}\\\dfrac{4}{5}.x+\dfrac{1}{3}=-\dfrac{2}{5}x-\dfrac{1}{3}\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}\left(\dfrac{4}{5}-\dfrac{2}{5}\right)x=\dfrac{1}{3}-\dfrac{1}{3}\\\left(\dfrac{4}{5}+\dfrac{2}{5}\right)x=-\dfrac{1}{3}-\dfrac{1}{3}\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}\dfrac{2}{5}x=0\\\dfrac{6}{5}x=-\dfrac{2}{3}\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=0:\dfrac{2}{5}\\x=-\dfrac{2}{3}:\dfrac{6}{5}\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=0\\-\dfrac{5}{9}\end{matrix}\right.\)
Vậy `....`
`#BaoL i n h`
Xét `2` trường hợp :
\(=>\left[{}\begin{matrix}\dfrac{4}{5}.x+\dfrac{1}{3}=\dfrac{2}{5}x+\dfrac{1}{3}\\\dfrac{4}{5}.x+\dfrac{1}{3}=-\dfrac{2}{5}x-\dfrac{1}{3}\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}\left(\dfrac{4}{5}-\dfrac{2}{5}\right)x=\dfrac{1}{3}-\dfrac{1}{3}\\\left(\dfrac{4}{5}+\dfrac{2}{5}\right)x=-\dfrac{1}{3}-\dfrac{1}{3}\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}\dfrac{2}{5}x=0\\\dfrac{6}{5}x=-\dfrac{2}{3}\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=0:\dfrac{2}{5}\\x=-\dfrac{2}{3}:\dfrac{6}{5}\end{matrix}\right.\)
`#LeMichael`
