a. Do M thuộc Ox, gọi tọa độ M có dạng \(M\left(a;0\right)\) \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{MA}=\left(-1-a;-2\right)\\\overrightarrow{MB}=\left(3-a;2\right)\\\overrightarrow{MC}=\left(4-a;-1\right)\end{matrix}\right.\)
\(\Rightarrow\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}=\left(6-3a;-1\right)\)
\(\Rightarrow T=\left|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right|=\sqrt{\left(6-3a\right)^2+1}\ge1\)
\(T_{min}=1\) khi \(6-3a=0\Leftrightarrow a=2\Rightarrow M\left(2;0\right)\)
b.
\(\overrightarrow{AB}=\left(4;4\right)\Rightarrow AB=\sqrt{4^2+4^2}=4\sqrt{2}\)
\(\overrightarrow{AC}=\left(5;1\right)\Rightarrow AC=\sqrt{5^2+1^2}=\sqrt{26}\)
\(\overrightarrow{BC}=\left(1;-3\right)\Rightarrow BC=\sqrt{1^2+\left(-3\right)^2}=\sqrt{10}\)
\(\Rightarrow AB+BC+AC=4\sqrt{2}+\sqrt{26}+\sqrt{10}\)
c.
Gọi M là trung điểm BC \(\Rightarrow M\left(\frac{7}{2};\frac{1}{2}\right)\)
\(\Rightarrow\overrightarrow{AM}=\left(\frac{9}{2};\frac{5}{2}\right)\Rightarrow AM=\sqrt{\left(\frac{9}{2}\right)^2+\left(\frac{5}{2}\right)^2}=\frac{\sqrt{106}}{2}\)
