17:
\(\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\)
\(\Rightarrow\left|\overrightarrow{AC}\right|=a\) vì tam giác ABC đều
18.
\(\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{BA}\)
\(\Rightarrow\left|\overrightarrow{ BA}\right|=2a\) vì tam giác ABC đều
19 .
\(\overrightarrow{CA}+\overrightarrow{AB}=\overrightarrow{CB}\)
\(\Rightarrow\left|\overrightarrow{CB}\right|=3a\) vì tam giác ABC đều
20.
\(\overrightarrow{BA}+\overrightarrow{AC}=\overrightarrow{BC}\)
\(\Rightarrow\left|\overrightarrow{BC}\right|=4a\) vì tam giác ABC đều
Mình chữa nốt 14, 15, 16, 21
14, \(\overrightarrow{BD}+\overrightarrow{CA}=\overrightarrow{BA}+\overrightarrow{CD}\)
\(\Leftrightarrow\) \(\overrightarrow{BD}-\overrightarrow{BA}=\overrightarrow{CD}-\overrightarrow{CA}\)
\(\Leftrightarrow\) \(\overrightarrow{AD}=\overrightarrow{AD}\) (luôn đúng)
15, \(\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CB}+\overrightarrow{DA}\)
\(\Leftrightarrow\) \(\overrightarrow{CA}-\overrightarrow{CB}=\overrightarrow{DA}-\overrightarrow{DB}\)
\(\Leftrightarrow\) \(\overrightarrow{BA}=\overrightarrow{BA}\) (luôn đúng)
16, \(\overrightarrow{DC}+\overrightarrow{AB}=\overrightarrow{DB}+\overrightarrow{AC}\)
\(\Leftrightarrow\) \(\overrightarrow{DC}-\overrightarrow{DB}=\overrightarrow{AC}-\overrightarrow{AB}\)
\(\Leftrightarrow\) \(\overrightarrow{BC}=\overrightarrow{BC}\) (luôn đúng)
21,
Ta có: \(\left|\overrightarrow{F}\right|=\left|\overrightarrow{F1}+\overrightarrow{F2}\right|\)
\(\Leftrightarrow\) \(F^2=F1^2+F2^2+2\overrightarrow{F1}\cdot\overrightarrow{F2}\)
\(\Leftrightarrow\) \(F^2=F1^2+F2^2+2F1\cdot F2\cdot cos\left(\overrightarrow{F1},\overrightarrow{F2}\right)\)
\(\Leftrightarrow\) \(F=\sqrt{500^2+500^2+2\cdot500\cdot500\cdot cos60^o}\)
\(\Leftrightarrow\) \(F\approx866\left(N\right)\)
Chúc bn học tốt!
nhìu quá cho bớt lại đc hơm , 20 21 thôi
