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AH
Akai Haruma
Giáo viên
15 tháng 7 2017

Lời giải:

Ta có:

\(\overrightarrow{AI}+\overrightarrow{FA}=\overrightarrow{AB}+\overrightarrow{BI}+\overrightarrow{FD}+\overrightarrow{DA}=\overrightarrow{AB}+\overrightarrow{DA}+\frac{1}{2}(\overrightarrow{BC}+\overrightarrow{CD})\)

Suy ra \(\text{VT}=4(\overrightarrow{AB}+\overrightarrow{DA})+\overrightarrow{BC}+\overrightarrow{CD}\)

\(\Leftrightarrow{VT}=3\overrightarrow{DB}+\overrightarrow{AB}+\overrightarrow{DA}+\overrightarrow{BC}+\overrightarrow{CD}\)

\(\Leftrightarrow{VT}=3\overrightarrow{DB}+(\overrightarrow{AB}+\overrightarrow{BC})+(\overrightarrow{CD}+\overrightarrow{DA})\)

\(\Leftrightarrow{VT}=3\overrightarrow{DB}+\overrightarrow{AC}+\overrightarrow{CA}=3\overrightarrow{DB}=\text{VP}\)

Ta có đpcm

17 tháng 8 2019

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 \(\Delta BCD\)

\(\Rightarrow\left\{{}\begin{matrix}IN//BD\\IN=\frac{1}{2}BD\end{matrix}\right.\Rightarrow\overrightarrow{IN}=\frac{1}{2}\overrightarrow{BD}\\ \Rightarrow2\left(\overrightarrow{AB}+\overrightarrow{AI}+\overrightarrow{NA}+\overrightarrow{DA}\right)\\ =2\left(\overrightarrow{DB}+\overrightarrow{NI}\right)=2\left(\overrightarrow{DB}+\frac{1}{2}\overrightarrow{DB}\right)=2\cdot\frac{3}{2}\overrightarrow{DB}=3\overrightarrow{DB}\)

29 tháng 9 2019

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

\(\Rightarrow\overrightarrow{IB}+\overrightarrow{IC}=2\overrightarrow{IF}=2\overrightarrow{HI}\)

\(\Rightarrow\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}=2\overrightarrow{HI}+\overrightarrow{ID}=\overrightarrow{HI}+\overrightarrow{HD}\)

\(\overrightarrow{HD}=\overrightarrow{AH}\Rightarrow\overrightarrow{IB}+\overrightarrow{IC}+\overrightarrow{ID}=\overrightarrow{HI}+\overrightarrow{AH}=\overrightarrow{AI}\)

Thay vào cái trên sẽ có đpcm

NV
23 tháng 2 2020

\(\overrightarrow{MA}+\overrightarrow{MC}=\overrightarrow{MB}+\overrightarrow{BA}+\overrightarrow{MD}+\overrightarrow{DC}=\overrightarrow{MB}+\overrightarrow{MD}+\overrightarrow{BA}+\overrightarrow{DC}=\overrightarrow{MB}+\overrightarrow{MD}\)

b/

\(2\left(\overrightarrow{JA}+\overrightarrow{AB}+\overrightarrow{DA}+\overrightarrow{AI}\right)=2\left(\overrightarrow{JB}+\overrightarrow{DI}\right)=2\left(\overrightarrow{JD}+\overrightarrow{DB}+\overrightarrow{DB}+\overrightarrow{BI}\right)\)

\(=2\left(2\overrightarrow{DB}+\overrightarrow{IC}+\overrightarrow{CJ}\right)=2\left(2\overrightarrow{DB}+\overrightarrow{IJ}\right)=2\left(2\overrightarrow{DB}+\frac{1}{2}\overrightarrow{BD}\right)=3\overrightarrow{DB}\)c/

\(\overrightarrow{AK}=\overrightarrow{AB}+\overrightarrow{BK}=\overrightarrow{AB}+\frac{1}{6}\overrightarrow{BD}=\overrightarrow{AB}+\frac{1}{6}\left(\overrightarrow{BA}+\overrightarrow{BC}\right)=\frac{5}{6}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{BC}\)

\(\overrightarrow{AH}=\overrightarrow{AB}+\overrightarrow{BH}=\overrightarrow{AB}+\frac{1}{5}\overrightarrow{BC}=\frac{6}{5}\left(\frac{5}{6}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{BC}\right)=\frac{6}{5}\overrightarrow{AK}\)

\(\Rightarrow A;K;H\) thẳng hàng

9 tháng 10 2019

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HQ
Hà Quang Minh
Giáo viên
24 tháng 9 2023

a)

 \(\begin{array}{l}\overrightarrow {AB}  + \overrightarrow {BC}  + \overrightarrow {CD}  + \overrightarrow {DA}  = \left( {\overrightarrow {AB}  + \overrightarrow {BC} } \right) + \left( {\overrightarrow {CD}  + \overrightarrow {DA} } \right)\\ = \overrightarrow {AC}  + \overrightarrow {CA}  = \overrightarrow {AA}  = \overrightarrow 0 .\end{array}\)

b)

\(\overrightarrow {AC}  - \overrightarrow {AD}  = \overrightarrow {DC} \) và \(\overrightarrow {BC}  - \overrightarrow {BD}  = \overrightarrow {DC} \)

\( \Rightarrow \overrightarrow {AC}  - \overrightarrow {AD}  = \overrightarrow {BC}  - \overrightarrow {BD} \)

HQ
Hà Quang Minh
Giáo viên
24 tháng 9 2023

Ta có: \(3\left( {\overrightarrow {AB}  + 2\overrightarrow {BC} } \right) - 2\left( {\overrightarrow {AB}  + 3\overrightarrow {BC} } \right)\)\( = 3\overrightarrow {AB}  + 3.\left( {2\overrightarrow {BC} } \right) - \left[ {2\overrightarrow {AB}  + 2.\left( {3\overrightarrow {BC} } \right)} \right]\)

\[ = 3\overrightarrow {AB}  + 6.\overrightarrow {BC}  - \left( {2\overrightarrow {AB}  + 6.\overrightarrow {BC} } \right)\]\[ = 3\overrightarrow {AB}  + 6.\overrightarrow {BC}  - 2\overrightarrow {AB}  - 6.\overrightarrow {BC} \]

\[ = \left( {3\overrightarrow {AB}  - 2\overrightarrow {AB} } \right) + \left( {6.\overrightarrow {BC}  - 6.\overrightarrow {BC} } \right) = \overrightarrow {AB} .\]