a)\(\overrightarrow{|AB}+\overrightarrow{OD|}=\left|\overrightarrow{AB}+\overrightarrow{BO}\right|=\left|\overrightarrow{AO}\right|=\frac{a\sqrt{2}}{2}\)

b)\(\left|\overrightarrow{MA}-\overrightarrow{MB}-\overrightarrow{MC}+\overrightarrow{MD}=\overrightarrow{BA}+\overrightarrow{CM}+\overrightarrow{MD}\right|=\left|\overrightarrow{BA}+\overrightarrow{CD}\right|=\left|\overrightarrow{BA}+\overrightarrow{BA}\right|=2a\)

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Ta có: \(AB = BC = CD = DA = 1;\)

            \(AC = BD = \sqrt {A{B^2} + B{C^2}}  = \sqrt {{1^2} + {1^2}}  = \sqrt 2 \)

a) \(\overrightarrow a  = \overrightarrow {OB}  - \overrightarrow {OD}  = \overrightarrow {OB}  + \overrightarrow {DO}  = \left( {\overrightarrow {DO}  + \overrightarrow {OB} } \right) = \overrightarrow {DB} \)

\( \Rightarrow \left| {\overrightarrow a } \right| = \left| {\overrightarrow {DB} } \right| = DB = \sqrt 2 \)

b)  \(\overrightarrow b = \left( {\overrightarrow {OC}  - \overrightarrow {OA} } \right) + \left( {\overrightarrow {DB}  - \overrightarrow {DC} } \right)\)

   \( = \left( {\overrightarrow {OC}  + \overrightarrow {AO} } \right) + \left( {\overrightarrow {DB}  + \overrightarrow {CD} } \right) = \left( {\overrightarrow {AO}  + \overrightarrow {OC} } \right) + \left( {\overrightarrow {CD}  + \overrightarrow {DB} } \right)\)

   \( = \overrightarrow {AC}  + \overrightarrow {CB}  = \overrightarrow {AB} \)

\( \Rightarrow \left| {\overrightarrow b } \right| = \left| {\overrightarrow {AB} } \right| = AB = 1\)

Chú ý khi giải:

Khi có dấu trừ phía trước ta thường thay bằng vectơ đối của nó và ngược lại

Gọi N là trung điểm BC

\(\left|\overrightarrow{MA}+\overrightarrow{MC}+2\overrightarrow{MB}+2\overrightarrow{OC}\right|=\left|\overrightarrow{AB}-\overrightarrow{AD}\right|\)

\(\Leftrightarrow\left|2\overrightarrow{MO}+2\overrightarrow{MB}+2\overrightarrow{OC}\right|=\left|\overrightarrow{AB}-\overrightarrow{AD}\right|\)

\(\Leftrightarrow\left|2\overrightarrow{MC}+2\overrightarrow{MB}\right|=\left|\overrightarrow{AB}-\overrightarrow{AD}\right|\)

\(\Leftrightarrow4\left|\overrightarrow{MN}\right|=\left|\overrightarrow{BD}\right|\)

\(\Rightarrow\left|\overrightarrow{BD}\right|=4\left|\overrightarrow{MN}\right|=4\left|\overrightarrow{DN}+\overrightarrow{MD}\right|\ge4MD-4DN\)

\(\Rightarrow4MD\le BD+4DN\)

\(\Leftrightarrow MD\le\dfrac{BD+4DN}{4}=\dfrac{a\sqrt{2}+2a\sqrt{5}}{4}=\dfrac{2\sqrt{5}+\sqrt{2}}{4}a\)

Bài 3: 

Tham khảo:

a/ \(\left|\overrightarrow{OA}+\overrightarrow{OC}+\overrightarrow{OB}+\overrightarrow{OD}\right|=\left|\overrightarrow{0}+\overrightarrow{0}\right|=0\)

b/ \(\left|\overrightarrow{OA}+\overrightarrow{OB}\right|+\left|\overrightarrow{OC}+\overrightarrow{OD}\right|=a+a=2a\)

c/

\(\left|\overrightarrow{OA}+\overrightarrow{OC}+\overrightarrow{OB}\right|+\left|\overrightarrow{OD}\right|=\left|\overrightarrow{OB}\right|+\left|\overrightarrow{OD}\right|=2\left|\overrightarrow{OB}\right|=2\sqrt{a^2-\frac{a^2}{4}}=a\sqrt{3}\)

\(\left|\overrightarrow{OA}-\overrightarrow{CB}\right|=\left|\overrightarrow{OA}+\overrightarrow{BC}\right|=\left|\overrightarrow{OA}+\overrightarrow{AD}\right|=\left|\overrightarrow{OD}\right|=OD=\dfrac{1}{2}BD=\dfrac{a\sqrt{2}}{2}\)

\(\left|\overrightarrow{AB}+\overrightarrow{DC}\right|=\left|\overrightarrow{AB}+\overrightarrow{AB}\right|=2\left|\overrightarrow{AB}\right|=2AB=2a\)

\(\left|\overrightarrow{CD}-\overrightarrow{DA}\right|=\left|\overrightarrow{CD}+\overrightarrow{AD}\right|=\left|\overrightarrow{BA}+\overrightarrow{AD}\right|=\left|\overrightarrow{BD}\right|=BD=a\sqrt{2}\)

Tham khảo:

Dễ thấy: \(\overrightarrow {OA}  = \overrightarrow {OM}  + \overrightarrow {MA} \); \(\overrightarrow {OB}  = \overrightarrow {OM}  + \overrightarrow {MB} \)

Tương tự: \(\overrightarrow {OC}  = \overrightarrow {ON}  + \overrightarrow {NC} \); \(\overrightarrow {OD}  = \overrightarrow {ON}  + \overrightarrow {ND} \)

\(\begin{array}{l} \Rightarrow \overrightarrow {OA}  + \overrightarrow {OB}  + \overrightarrow {OC}  + \overrightarrow {OD}  = \left( {\overrightarrow {OM}  + \overrightarrow {MA} } \right) + \left( {\overrightarrow {OM}  + \overrightarrow {MB} } \right) + \left( {\overrightarrow {ON}  + \overrightarrow {NC} } \right) + \left( {\overrightarrow {ON}  + \overrightarrow {ND} } \right)\\ = \left( {\overrightarrow {OM}  + \overrightarrow {OM}  + \overrightarrow {MA}  + \overrightarrow {MB} } \right) + \left( {\overrightarrow {ON}  + \overrightarrow {ON}  + \overrightarrow {NC}  + \overrightarrow {ND} } \right)\\ = \overrightarrow {OM}  + \overrightarrow {OM}  + \overrightarrow {ON}  + \overrightarrow {ON} \\ = \left( {\overrightarrow {OM}  + \overrightarrow {ON} } \right) + \left( {\overrightarrow {OM}  + \overrightarrow {ON} } \right)\\ = \overrightarrow 0  + \overrightarrow 0 \\ = \overrightarrow 0 .\end{array}\)