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a/ \(VT=\overrightarrow{AB}+\overrightarrow{BF}+\overrightarrow{BC}+\overrightarrow{CG}+\overrightarrow{CD}+\overrightarrow{DH}+\overrightarrow{DA}+\overrightarrow{AE}\)
\(=\left(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}\right)+\left(\frac{1}{2}\overrightarrow{BC}+\frac{1}{2}\overrightarrow{CD}+\frac{1}{2}\overrightarrow{DA}+\frac{1}{2}\overrightarrow{AB}\right)\)
\(=\overrightarrow{0}+\frac{1}{2}.\overrightarrow{0}=\overrightarrow{0}=VP\)
b/ Câu này áp dụng luôn kq câu a
\(\overrightarrow{MF}-\overrightarrow{MA}+\overrightarrow{MG}-\overrightarrow{MB}+\overrightarrow{MH}-\overrightarrow{MC}+\overrightarrow{ME}-\overrightarrow{MD}=\overrightarrow{0}\)
chuyển mấy cái vecto kia sang vế phải là có ngay đpcm câu b
c/\(VT=\overrightarrow{AI}+\overrightarrow{IB}+\overrightarrow{AI}+\overrightarrow{IC}+\overrightarrow{AI}+\overrightarrow{ID}=3\overrightarrow{AI}+\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}\)
Để ý tới G là TĐ CD, F là TĐ BC
Theo quy tắc trung điểm
\(\Rightarrow\overrightarrow{IB}+\overrightarrow{IC}=2\overrightarrow{IF}=2\overrightarrow{HI}\)
\(\Rightarrow\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}=2\overrightarrow{HI}+\overrightarrow{ID}=\overrightarrow{HI}+\overrightarrow{HD}\)
Mà \(\overrightarrow{HD}=\overrightarrow{AH}\Rightarrow\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}=\overrightarrow{HI}+\overrightarrow{AH}=\overrightarrow{AI}\)
Thay vào cái trên sẽ có đpcm
\(\begin{array}{l}\overrightarrow {MA} + \overrightarrow {MB} + \overrightarrow {MC} + \overrightarrow {MD} = \left( {\overrightarrow {MG} + \overrightarrow {GE} + \overrightarrow {EA} } \right) + \left( {\overrightarrow {MG} + \overrightarrow {GE} + \overrightarrow {EB} } \right)\\ + \left( {\overrightarrow {MG} + \overrightarrow {GF} + \overrightarrow {FC} } \right) + \left( {\overrightarrow {MG} + \overrightarrow {GF} + \overrightarrow {FD} } \right)\end{array}\)
\( = \left( {\overrightarrow {MG} + \overrightarrow {MG} + \overrightarrow {MG} \overrightarrow { + MG} } \right) + 2\left( {\overrightarrow {GE} + \overrightarrow {GF} } \right) \\+ \left( {\overrightarrow {EA} + \overrightarrow {EB} } \right) + \left( {\overrightarrow {FC} + \overrightarrow {FD} } \right)\)
\( = 4\overrightarrow {MG} + 2.\overrightarrow 0 + \overrightarrow 0 + \overrightarrow 0 = 4\overrightarrow {MG} \) (đpcm)
Chắc chắn là đề bài sai rồi
Vế trái là 1 đại lượng vô hướng
Vế phải là 1 đại lượng có hướng (vecto)
Hai vế không thể bằng nhau được
\(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AB}+\overrightarrow{CB}+\overrightarrow{BD}=\overrightarrow{AB}+\overrightarrow{BD}+\overrightarrow{CB}=\overrightarrow{AD}+\overrightarrow{CB}\)
\(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=\left(\overrightarrow{OE}+\overrightarrow{EA}\right)+\left(\overrightarrow{OF}+\overrightarrow{FB}\right)+\left(\overrightarrow{OE}+\overrightarrow{EC}\right)+\left(\overrightarrow{OF}+\overrightarrow{FD}\right)\)
\(=2\left(\overrightarrow{OE}+\overrightarrow{EF}\right)+\left(\overrightarrow{EA}+\overrightarrow{EC}\right)+\left(\overrightarrow{FB}+\overrightarrow{FD}\right)\)
\(=2.\overrightarrow{0}+\overrightarrow{0}+\overrightarrow{0}=\overrightarrow{0}\)
a) Chữa đề: \(\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CB}+\overrightarrow{DA}=2\overrightarrow{NM}\)
\(Ta\text{ }có:\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CB}+\overrightarrow{BA}+\overrightarrow{DA}+\overrightarrow{AB}\\ =\overrightarrow{CB}+\overrightarrow{DA}+\left(\overrightarrow{BA}+\overrightarrow{AB}\right)=\overrightarrow{CB}+\overrightarrow{DA}\)
\(\)\(\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CA}+\overrightarrow{CB}+\overrightarrow{DC}\\ =2\overrightarrow{CM}+2\overrightarrow{NC}=2\left(\overrightarrow{NC}+\overrightarrow{CM}\right)=2\overrightarrow{NM}\)
Vậy \(\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CB}+\overrightarrow{DA}=2\overrightarrow{NM}\)
\(\text{b) }\overrightarrow{AD}+\overrightarrow{BD}+\overrightarrow{AC}+\overrightarrow{BC}=-\left(\overrightarrow{DA}+\overrightarrow{DB}+\overrightarrow{CA}+\overrightarrow{CB}\right)\\ =-\left[\left(\overrightarrow{DA}+\overrightarrow{DB}\right)+\left(\overrightarrow{CA}+\overrightarrow{CB}\right)\right]\\ =-\left(2\overrightarrow{DM}+2\overrightarrow{CM}\right)=2\left(\overrightarrow{MD}+\overrightarrow{MC}\right)=4\left(\overrightarrow{MN}\right)\)
\(\text{c) }2\left(\overrightarrow{AB}+\overrightarrow{AI}+\overrightarrow{NA}+\overrightarrow{DA}\right)\\ =2\left[\left(\overrightarrow{AB}+\overrightarrow{DA}\right)+\left(\overrightarrow{AI}+\overrightarrow{NA}\right)\right]\\ =2\left[\left(\overrightarrow{AB}+\overrightarrow{BA}+\overrightarrow{DB}\right)+\overrightarrow{NI}\right]=2\left(\overrightarrow{DB}+\overrightarrow{NI}\right)\)
Mà IN là dường trung bình \(\Delta BCD\)
\(\Rightarrow\left\{{}\begin{matrix}IN//BD\\IN=\frac{1}{2}BD\end{matrix}\right.\Rightarrow\overrightarrow{IN}=\frac{1}{2}\overrightarrow{BD}\\ \Rightarrow2\left(\overrightarrow{AB}+\overrightarrow{AI}+\overrightarrow{NA}+\overrightarrow{DA}\right)\\ =2\left(\overrightarrow{DB}+\overrightarrow{NI}\right)=2\left(\overrightarrow{DB}+\frac{1}{2}\overrightarrow{DB}\right)=2\cdot\frac{3}{2}\overrightarrow{DB}=3\overrightarrow{DB}\)
a: \(\overrightarrow{AB}+\overrightarrow{DC}=\overrightarrow{AI}+\overrightarrow{IB}+\overrightarrow{DI}+\overrightarrow{IC}\)
\(=\overrightarrow{AI}+\overrightarrow{DI}=-\left(\overrightarrow{IA}+\overrightarrow{ID}\right)=-2\overrightarrow{IM}=2\overrightarrow{MI}\)
\(\overrightarrow{AB}+\overrightarrow{DC}=\overrightarrow{AC}+\overrightarrow{DB}\)
\(\Leftrightarrow\overrightarrow{AB}-\overrightarrow{AC}=\overrightarrow{DB}-\overrightarrow{DC}\)
\(\Leftrightarrow\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{CD}+\overrightarrow{DB}=\overrightarrow{CB}\)(luôn đúng)
=>ĐPCM
b: \(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD}\)
\(=2\cdot\overrightarrow{GM}+2\cdot\overrightarrow{GI}=\overrightarrow{0}\)
a) \(\overrightarrow {AC} + \overrightarrow {BD} = \overrightarrow {AM} + \overrightarrow {MN} + \overrightarrow {NC} + \overrightarrow {BM} + \overrightarrow {MN} + \overrightarrow {ND} \\= \left( {\overrightarrow {AM} + \overrightarrow {BM} } \right) + \left( {\overrightarrow {MN} + \overrightarrow {MN} } \right) + \left( {\overrightarrow {NC} + \overrightarrow {ND} } \right) \\= \overrightarrow 0 + 2\overrightarrow {MN} + \overrightarrow 0 = 2\overrightarrow {MN} \) (đpcm)
b) \(\overrightarrow {AC} + \overrightarrow {BD} = \overrightarrow {BC} + \overrightarrow {AD} \)
\(\)\(\overrightarrow {BC} + \overrightarrow {AD} = \overrightarrow {BM} + \overrightarrow {MN} + \overrightarrow {NC} + \overrightarrow {AM} + \overrightarrow {MN} + \overrightarrow {ND} \)
\(\left( {\overrightarrow {BM} + \overrightarrow {AM} } \right) + \left( {\overrightarrow {MN} + \overrightarrow {MN} } \right) + \left( {\overrightarrow {NC} + \overrightarrow {ND} } \right) = 2\overrightarrow {MN} \)
Mặt khác ta có: \(\overrightarrow {AC} + \overrightarrow {BD} = 2\overrightarrow {MN} \)
Suy ra \(\overrightarrow {AC} + \overrightarrow {BD} = \overrightarrow {BC} + \overrightarrow {AD} \)
Cách 2:
\(\begin{array}{l}
\overrightarrow {AC} + \overrightarrow {BD} = \overrightarrow {BC} + \overrightarrow {AD} \\
\Leftrightarrow \overrightarrow {AC} - \overrightarrow {AD} = \overrightarrow {BC} - \overrightarrow {BD} \\
\Leftrightarrow \overrightarrow {DC} = \overrightarrow {DC} (đpcm)
\end{array}\)
