\(AB=\sqrt{\left(2+1\right)^2+\left(3-2\right)^2}=\sqrt{10}\)
\(AC=\sqrt{\left(m+1\right)^2+\left(0-2\right)^2}=\sqrt{\left(m+1\right)^2+4}\ge2,\forall m\in R\)
\(BC\sqrt{\left(m-2\right)^2+\left(0-3\right)^2}=\sqrt{\left(m-2\right)^2+9}\ge3,\forall x\in m\)
Chu vi tam giác ABC :
\(P=AB+AC+BC=\sqrt{\left(m+1\right)^2+4}+\sqrt{\sqrt{\left(m-2\right)^2+9}}+\sqrt{10}\)
Nếu \(AC=2\Leftrightarrow m+1=0\Leftrightarrow m=-1\)
\(\Rightarrow P=2+3\sqrt{2}+\sqrt{10}\left(1\right)\)
\(\Rightarrow P\ge12+3\sqrt{2}\left(1\right)\)
Nếu \(BC=3\Leftrightarrow m-2=0\Leftrightarrow m=2\)
\(P=\sqrt{13}+3+\sqrt{10}\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow P\left(min\right)=2+3\sqrt{2}+\sqrt{10}\) tại \(m=-1\)
