Cho hình vuông $ABCD$, khẳng định nào sau đây đúng:

$a\sqrt 3$.

$\dfrac{{a\sqrt 3 }}{2}$.

$2a$.

$a$.

$m > \dfrac{3}{2}$.

$m > \dfrac{3}{4}$.

$\dfrac{3}{4} < m < \dfrac{3}{2}$.

$1 < m < 3$.

$2S$

$3S$

$4S$   

$6S$

$- \dfrac{1}{2}x - y + 1 \le 0$

$- \dfrac{1}{2}x - y + 1 > 0$

$- \dfrac{1}{2}x - y + 1 \ge 0$

$- \dfrac{1}{2}x - y + 1 < 0$

$y = 2{\left( {x + 3} \right)^2}$.

$y = 2{x^2} + 3$

$y = 2{\left( {x - 3} \right)^2}$.

$y = 2{x^2} - 3$.

$\left( {\overrightarrow {AB} .\overrightarrow {AC} } \right)\overrightarrow {BC}  = 2\overrightarrow {BC}$.

$\overrightarrow {BC} .\overrightarrow {CA} =  - 2$.

$\left( {\overrightarrow {AB}  + \overrightarrow {BC} } \right).\overrightarrow {AC}  =  - 4$.

$\left( {\overrightarrow {BC}  - \overrightarrow {AC} } \right).\overrightarrow {BA}  = 4$.