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\(\overrightarrow{NP}=\overrightarrow{NC}+\overrightarrow{CP}\)
\(=\dfrac{2}{3}\overrightarrow{BC}+\dfrac{1}{3}\overrightarrow{CA}\)
\(=-\dfrac{2}{3}\overrightarrow{CB}+\dfrac{1}{3}\overrightarrow{CA}\)
\(\overrightarrow{PM}=\overrightarrow{PA}+\overrightarrow{AM}\)
\(=\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{AB}\)
\(=\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\left(\overrightarrow{AC}+\overrightarrow{CB}\right)\)
\(=\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{CB}\)
1.
Gọi M là trung điểm BC thì theo tính chất trọng tâm: \(\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AM}=\dfrac{2}{3}\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)\)
\(\Rightarrow\overrightarrow{AG}=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\Rightarrow x+y=\dfrac{2}{3}\)
2.
\(CH=\dfrac{1}{2}BC=\dfrac{a}{2}\)
\(T=\left|\text{ }\overrightarrow{CA}-\overrightarrow{HC}\right|=\left|\overrightarrow{CA}+\overrightarrow{CH}\right|\)
\(\Rightarrow T^2=CA^2+CH^2+2\overrightarrow{CA}.\overrightarrow{CH}=a^2+\left(\dfrac{a}{2}\right)^2+2.a.\dfrac{a}{2}.cos60^0=\dfrac{7a^2}{4}\)
\(\Rightarrow T=\dfrac{a\sqrt{7}}{2}\)
3.
\(10< x< 100\Rightarrow10< 3k< 100\)
\(\Rightarrow\dfrac{10}{3}< k< \dfrac{100}{3}\Rightarrow4\le k\le33\)
\(\Rightarrow\sum x=3\left(4+5+...+33\right)=1665\)
1) Ta có:\(\overrightarrow{AB}+\overrightarrow{DE}-\overrightarrow{DB}+\overrightarrow{BC}=\overrightarrow{AE}+\overrightarrow{BC}=\overrightarrow{AC}+\overrightarrow{CE}+\overrightarrow{BE}+\overrightarrow{EC}\)
\(=\overrightarrow{AC}+\overrightarrow{BE}+\overrightarrow{CE}+\overrightarrow{EC}=\overrightarrow{AC}+\overrightarrow{BE}\left(đpcm\right)\)2) a) Ta có: \(\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}=\overrightarrow{AE}+\overrightarrow{ED}+\overrightarrow{BF}+\overrightarrow{FE}+\overrightarrow{CD}+\overrightarrow{DF}\)\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}+\overrightarrow{ED}+\overrightarrow{DF}+\overrightarrow{FE}\)
\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}\left(đpcm\right)\)
b) Ta có: \(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AD}+\overrightarrow{DB}+\overrightarrow{CB}+\overrightarrow{BD}\)
\(=\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{DB}+\overrightarrow{BD}=\overrightarrow{AD}+\overrightarrow{CB}\left(đpcm\right)\)c) \(\overrightarrow{AB}-\overrightarrow{CD}=\overrightarrow{AB}-\overrightarrow{BD}\)
\(\overrightarrow{AB}+\overrightarrow{DC}=\overrightarrow{AB}+\overrightarrow{DB}\)
Ta có: \(\overrightarrow{AB}+\overrightarrow{DC}=\overrightarrow{AB}+\overrightarrow{DB}+\overrightarrow{BC}\) ( đề bài bị lỗi gì à ?? :v ) hay do mình =))
Do G là trọng tâm tam giác
\(\Rightarrow\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AD}=\dfrac{2}{3}\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}=\dfrac{1}{3}\overrightarrow{AC}+\dfrac{1}{3}\overrightarrow{CB}+\dfrac{1}{3}\overrightarrow{AC}\)
\(=\dfrac{2}{3}\overrightarrow{AC}+\dfrac{1}{3}\overrightarrow{CB}=-\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{CB}\)
Do I là trung điểm AG
\(\Rightarrow\overrightarrow{AI}=\dfrac{1}{2}\overrightarrow{AG}=\dfrac{1}{2}\left(-\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{CB}\right)=-\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{6}\overrightarrow{CB}\)
\(\overrightarrow{AK}=\dfrac{1}{5}\overrightarrow{AB}=\dfrac{1}{5}\left(\overrightarrow{AC}+\overrightarrow{CB}\right)=-\dfrac{1}{5}\overrightarrow{CA}+\dfrac{1}{5}\overrightarrow{CB}\)
\(\overrightarrow{CI}=\overrightarrow{CA}+\overrightarrow{AI}=\overrightarrow{CA}-\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{6}\overrightarrow{CB}=\dfrac{2}{3}\overrightarrow{CA}+\dfrac{1}{6}\overrightarrow{CB}\)
\(\overrightarrow{CK}=\overrightarrow{CA}+\overrightarrow{AK}=\overrightarrow{CA}-\dfrac{1}{5}\overrightarrow{CA}+\dfrac{1}{5}\overrightarrow{CB}=\dfrac{4}{5}\overrightarrow{CA}+\dfrac{1}{5}\overrightarrow{CB}\)
\(a,\overrightarrow{AB}=\left(2;10\right)\)
\(\overrightarrow{AC}=\left(-5;5\right)\)
\(\overrightarrow{BC}=\left(-7;-5\right)\)
\(b,\) Thiếu dữ kiện
\(c,Cos\left(\overrightarrow{AB},\overrightarrow{AC}\right)=\dfrac{\left|2\left(-5\right)+10.5\right|}{\sqrt{2^2+10^2}.\sqrt{\left(-5\right)^2+5^2}}=\dfrac{2\sqrt{13}}{13}\)
\(\Rightarrow\left(\overrightarrow{AB},\overrightarrow{AC}\right)=56^o18'\)
\(Cos\left(\overrightarrow{AB},\overrightarrow{BC}\right)=\dfrac{\left|2\left(-7\right)+10\left(-5\right)\right|}{\sqrt{2^2+10^2}.\sqrt{\left(-7\right)^2+\left(-5\right)^2}}\)
\(\Rightarrow\left(\overrightarrow{AB},\overrightarrow{BC}\right)=43^o9'\)
bài 1
a CO-OB=BA
<=.> CO = BA +OB
<=> CO=OA ( LUÔN ĐÚNG )=>ĐPCM
b AB-BC=DB
<=> AB=DB+BC
<=> AB=DC(LUÔN ĐÚNG )=> ĐPCM
Cc DA-DB=OD-OC
<=> DA+BD= OD+CO
<=> BA= CD (LUÔN ĐÚNG )=> ĐPCM
d DA-DB+DC=0
VT= DA +BD+DC
= BA+DC
Mà BA=CD(CMT)
=> VT= CD+DC=O