Bài 3 bạn tự vẽ hình nhá
[tex]\widehat{BAD}=60^{\circ}=>DO=2AO[/tex]
[tex]\left | \underset{AB}{\rightarrow}+\underset{AD}{\rightarrow} \right |=\left | \underset{AC}{\rightarrow} \right |[/tex]
[tex]=>\left | \underset{AC}{\rightarrow} \right |=AC=2AO=\frac{4a}{\sqrt{5}}[/tex](Pi-ta-go)
[tex]\left | \underset{BA}{\rightarrow}-\underset{BC}{\rightarrow} \right |=\left | \underset{CA}{\rightarrow} \right |=CA=\frac{4a}{\sqrt{5}}[/tex]
[tex]\left | \underset{OB}{\rightarrow}-\underset{DC}{\rightarrow} \right |=\left | \underset{DO}{\rightarrow}-\underset{DC}{\rightarrow} \right |=\left | \underset{CO}{\rightarrow} \right |=CO=\frac{1}{2}AC=\frac{2a}{\sqrt{5}}[/tex]
Bài 4
[tex]\left | \underset{OA}{\rightarrow}-\underset{CB}{\rightarrow} \right |=\left | \underset{CO}{\rightarrow}-\underset{CB}{\rightarrow} \right |=\left | \underset{BO}{\rightarrow} \right |=BO=\frac{a}{\sqrt{2}}[/tex](PI-ta-go)
[tex]\left | \underset{AB}{\rightarrow}+\underset{DC}{\rightarrow} \right |=\left | 2\underset{AB}{\rightarrow} \right |=2AB=2a[/tex]
[tex]\left | \underset{CD}{\rightarrow}-\underset{DA}{\rightarrow} \right |=\left | \underset{CD}{\rightarrow}+\underset{AD}{\rightarrow} \right |=\left | \underset{BA}{\rightarrow}+\underset{AD}{\rightarrow} \right |=\left | \underset{BD}{\rightarrow} \right |=\sqrt{2}a[/tex](PI-ta-go)
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