|x-1|=\(\frac{3}{5}\)
\(\Rightarrow x-1=\pm\frac{3}{5}\)
\(Th1:x-1=\frac{3}{5}\)
\(\Rightarrow x=\frac{8}{5}\)
\(Th2:x-1=-\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
|x-1|=3/5
=>x-1=±3/5
Th1:x-1=3/5
=>x=8/5
Th2:x-1=-3/5
=>x=2/5
/x-1/ \(=\frac{3}{5}\)
\(\Rightarrow\int^{x-1=\frac{3}{5}}_{x-1=-\frac{3}{5}}\)
\(\Rightarrow\int^{x=\frac{3}{5}+1}_{x=-\frac{3}{5}+1}\)
\(\Rightarrow\int^{x=\frac{8}{5}}_{x=\frac{2}{5}}\)
Vậy \(x=\frac{2}{5},\frac{8}{5}\)
Nhớ ấn đúng ạ
\(\left|x-1\right|=\frac{3}{5}\)
\(\Rightarrow\left|x-1\right|=\)+\(\frac{3}{5}\)
TH1 : \(x-1=\frac{3}{5}\)
\(x=\frac{3}{5}+1\)
\(x=\frac{8}{5}\)
TH2 :\(x-1=-\frac{3}{5}\)
\(x=\left(-\frac{3}{5}\right)+1\)
\(x=\frac{2}{5}\)
Vậy :\(x=\frac{8}{5};\frac{2}{5}\)
