\(\Leftrightarrow\left(3-m\right)^2=2\left|m-1\right|\)
\(\Leftrightarrow9-6m+m^2=2\left|m-1\right|\left(1\right)\)
TH1: \(m>1\)
\(\left(1\right)\Leftrightarrow9-6m+m^2=2m-2\)
\(\Leftrightarrow m^2-8m+11=0\)
\(\Leftrightarrow\left(m-4\right)^2=5\)
\(\Leftrightarrow\left[{}\begin{matrix}m=4+\sqrt{5}\left(tm\right)\\m=4-\sqrt{5}\left(tm\right)\end{matrix}\right.\)
TH2: \(m< 1\)
\(\left(1\right)\Leftrightarrow9-6m+m^2=2-2m\)
\(\Leftrightarrow m^2-4m+7=0\)
\(\Leftrightarrow\left(m-2\right)^2=-3\)
\(\Rightarrow\text{vô nghiệm}\)
Vậy pt đã cho có 2 nghiệm ...
