Question

\(\sqrt{\left(1-x\right)^2}=2x+1\)

Answer 1

\(\sqrt{\left(1-x\right)^2}=2x+1\)

\(\Leftrightarrow\left|1-x\right|=2x+1\)

TH1: \(\left|1-x\right|=1-x\) với \(1-x\ge0\Leftrightarrow x\le1\) 

Pt trở thành:

\(1-x=2x+1\) (ĐK: \(x\le1\)

\(\Leftrightarrow3x=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

TH2: \(\left|1-x\right|=-\left(1-x\right)\) với \(1-x< 0\Leftrightarrow x>1\)

Pt trở thành: 

\(-\left(1-x\right)=2x+1\) (ĐK: \(x>1\))

\(\Leftrightarrow-1+x=2x+1\)

\(\Leftrightarrow x=2\left(tm\right)\)

Vậy: \(S=\left\{0;2\right\}\)