\(\left|x\right|-x=\frac{2}{3}\)
\(\left|x\right|=\frac{2}{3}+x\)
\(x=\pm\left(\frac{2}{3}+x\right)\)
TH1:
\(x=\frac{2}{3}+x\)
\(x-x=\frac{2}{3}\)
\(0=\frac{2}{3}\)
=> loại
TH2:
\(x=-\frac{2}{3}-x\)
\(x+x=-\frac{2}{3}\)
\(2x=-\frac{2}{3}\)
\(x=-\frac{2}{3}\div2\)
\(x=-\frac{2}{3}\times\frac{1}{2}\)
\(x=-\frac{1}{3}\)
Vậy \(x=-\frac{1}{3}\)
TH1:x<0
Ta có:|x|-x=\(\frac{2}{3}\)
x-(-x)=\(\frac{2}{3}\)
x+x=\(\frac{2}{3}\)
2x=\(\frac{2}{3}\)
x=\(\frac{1}{3}\)
Vậy x=\(\frac{1}{3}\)
TH2:x lớn hơn 0
Ta có|x|-x=\(\frac{2}{3}\)
x-x=2/3
Suy ra x=\(\frac{2}{3}\) hoặc -\(\frac{2}{3}\)