Cho tam giác vuông cân ABC cạnh huyền bằng a. khi đó giá trị của biểu thức

* Hướng dẫn giải

$\begin{array}{l}{\left(\overrightarrow{AB}+\text{}\overrightarrow{BC}+\text{}\overrightarrow{CA}\right)}^2={\overrightarrow{AB}}^2+\text{}{\overrightarrow{BC}}^2+\text{}{\overrightarrow{CA}}^2+\text{}2.(\overrightarrow{AB}.\text{\hspace{0.17em}\hspace{0.17em}}\overrightarrow{BC}+\text{}\overrightarrow{BC}.\text{\hspace{0.17em}\hspace{0.17em}}\overrightarrow{CA}+\text{}\overrightarrow{CA}.\text{}\overrightarrow{AB})\\ \iff 2.(\overrightarrow{AB}.\text{\hspace{0.17em}\hspace{0.17em}}\overrightarrow{BC}+\text{}\overrightarrow{BC}.\text{\hspace{0.17em}\hspace{0.17em}}\overrightarrow{CA}+\text{}\overrightarrow{CA}.\text{}\overrightarrow{AB})\\ ={\left(\overrightarrow{AB}+\text{}\overrightarrow{BC}+\text{}\overrightarrow{CA}\right)}^2=-{\overrightarrow{AB}}^2-\text{}{\overrightarrow{BC}}^2-\text{}\overrightarrow{CA}\\ ={\overrightarrow{0}}^2-A{B}^2-B{C}^2-C{A}^2=0-a^2-a^2=-2a^2\\ \iff \overrightarrow{AB}.\text{\hspace{0.17em}\hspace{0.17em}}\overrightarrow{BC}+\text{}\overrightarrow{BC}.\text{\hspace{0.17em}\hspace{0.17em}}\overrightarrow{CA}+\text{}\overrightarrow{CA}.\text{}\overrightarrow{AB}=-a^2\end{array}$

Đáp án B