Câu 2:

$\overrightarrow{IA} + 3\overrightarrow{IB} = \overrightarrow{0}$

$3MA^2 + MB^2 = 3(\overrightarrow{MI} + \overrightarrow{IA})^2 + (\overrightarrow{MI} + \overrightarrow{IB})^2$

$ = 3MI^2 + 6\overrightarrow{MI}.\overrightarrow{IA} + 3IA^2 + MI^2 + 2\overrightarrow{MI}.\overrightarrow{IB} + IB^2$

$= 4MI^2 + 2\overrightarrow{MI}(3\overrightarrow{IA} + \overrightarrow{IB}) + 3IA^2 + IB^2$

$ = 4MI^2 + 2\overrightarrow{MI}.(-\overrightarrow{IA}) + 3IA^2 + IB^2$

$= 4MI^2 + 3IA^2 + IB^2 - 2\overrightarrow{MI}.\overrightarrow{IA}$

Mà $\overrightarrow{IA} + 3\overrightarrow{IB} = \overrightarrow{0} \Leftrightarrow \overrightarrow{AI} = 3\overrightarrow{IB} \Leftrightarrow \overrightarrow{IA} = -3\overrightarrow{IB}$

$\Rightarrow \overrightarrow{IA} = \frac{3}{4}\overrightarrow{BA}$

$\Leftrightarrow IA = \frac{3}{4}BA = 3a$

$\Rightarrow IB = \frac{1}{4}AB = a$

$3IA^2 + IB^2 = 3(3a)^2 + a^2 = 28a^2$

$3MA^2 + MB^2 = 4MI^2 + 28a^2 - 2\overrightarrow{MI}.\overrightarrow{IA}$

Để $3MA^2 + MB^2$ nhỏ nhất thì $\overrightarrow{MI} = \overrightarrow{0} \Leftrightarrow M \equiv I$

Vậy `\min (3MA^2 + MB^2) = 28a^2`

`->` ko có đáp án đúng