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a: \(\overrightarrow{AM}+\overrightarrow{BN}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{BC}=\dfrac{1}{2}\overrightarrow{AC}\)
b: \(=\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{BA}\)
\(=\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{BA}\)
c: \(\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}\)
\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{BC}+\dfrac{1}{2}\overrightarrow{CA}\)
\(=\dfrac{1}{2}\left(\overrightarrow{AC}+\overrightarrow{CA}\right)=\overrightarrow{0}\)
1.
Gọi G là trọng tâm tam giác
\(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{0}\)
\(\Leftrightarrow3\overrightarrow{OG}=\overrightarrow{0}\)
\(\Leftrightarrow O\equiv G\)
\(\Rightarrow O\) là trọng tâm tam giác ABC
\(\Rightarrow\Delta ABC\) đều
Gọi độ dài các cạnh tam giác là a
\(\overrightarrow{BN}.\overrightarrow{AM}=\dfrac{1}{4}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\left(\overrightarrow{BA}+\overrightarrow{BC}\right)=-\dfrac{1}{4}a^2-\dfrac{1}{8}a^2-\dfrac{1}{8}a^2+\dfrac{1}{2}a^2=0\)
Mặt khác \(\overrightarrow{BN}.\overrightarrow{AM}=BN.AM.cos\left(\overrightarrow{AM};\overrightarrow{BN}\right)\)
\(\Rightarrow BN.AM.cos\left(\overrightarrow{AM};\overrightarrow{BN}\right)=0\Rightarrow cos\left(\overrightarrow{AM};\overrightarrow{BN}\right)=0\Rightarrow\left(\overrightarrow{AM};\overrightarrow{BN}\right)=90^o\)
\(BD=\dfrac{AB}{cos45^o}=\dfrac{a}{\dfrac{\sqrt{2}}{2}}=a\sqrt{2}\)
\(\overrightarrow{BQ}.\overrightarrow{BP}=\dfrac{1}{4}\left(\overrightarrow{BA}+\overrightarrow{BD}\right)\left(\overrightarrow{BC}+\overrightarrow{BD}\right)\)
\(=\dfrac{1}{4}BA.BC.cos90^o+\dfrac{1}{4}BA.BD.cos45^o+\dfrac{1}{4}BD.BC.cos45^o+\dfrac{1}{4}BD^2\)
\(=\dfrac{1}{4}a^2+\dfrac{1}{4}a^2+\dfrac{1}{2}a^2=a^2\)
b)
\(\overrightarrow{AN}=\dfrac{1}{2}\overrightarrow{AO}=-\dfrac{1}{2}\overrightarrow{OA}\)
Vậy \(m=-\dfrac{1}{2};n=0\).
c)
\(\overrightarrow{MN}=\dfrac{1}{2}\overrightarrow{AB}=\dfrac{1}{2}\left(\overrightarrow{AO}+\overrightarrow{OB}\right)=-\dfrac{1}{2}\overrightarrow{OA}+\dfrac{1}{2}\overrightarrow{OB}\).
Vậy \(m=-\dfrac{1}{2};n=\dfrac{1}{2}\).
d)
\(\overrightarrow{MB}=\dfrac{1}{2}\overrightarrow{OB}\)
Vậy \(m=0;n=\dfrac{1}{2}\).
a)Ta có:
\(\overrightarrow{OA}+\overrightarrow{OM}+\overrightarrow{ON}=\overrightarrow{CO}+\dfrac{1}{2}\left(\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OC}+\overrightarrow{OD}\right)\)
\(=\overrightarrow{CO}+\dfrac{1}{2}.2\overrightarrow{OC}\)
\(=\overrightarrow{0}\)
\(\RightarrowĐPCM\)
b) Ta có:
\(\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AD}+2\overrightarrow{AB}\right)\)
\(\Rightarrow2\overrightarrow{AM}=\overrightarrow{AD}+2\overrightarrow{AB}\) (1)
Mà \(2\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{AC}\)(2)
Từ (1)(2) =>\(\overrightarrow{AD}+2\overrightarrow{AB}=\overrightarrow{AB}+\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{AC}+\overrightarrow{AB}=\overrightarrow{AB}+\overrightarrow{AC}\)
\(\RightarrowĐPCM\)
\(3\overrightarrow{AC}=2\overrightarrow{CB}\leftrightarrow3\overrightarrow{AO}+3\overrightarrow{OC}=2\overrightarrow{CO}+2\overrightarrow{OB}\)
\(\rightarrow5\overrightarrow{OC}=3\overrightarrow{OA}+2\overrightarrow{OB}\)
\(\rightarrow\overrightarrow{OC}=\dfrac{3}{5}\overrightarrow{OA}+\dfrac{2}{5}\overrightarrow{OB}\)
Ta có:
\(\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AO}+\overrightarrow{AB}\right)\)
\(\Leftrightarrow\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AO}+\overrightarrow{AO}+\overrightarrow{OB}\right)\)
\(\Leftrightarrow\overrightarrow{AM}=\dfrac{1}{2}\left(2\overrightarrow{AO}+\overrightarrow{OB}\right)\)
\(\Leftrightarrow\overrightarrow{AM}=\overrightarrow{AO}+\dfrac{1}{2}\overrightarrow{OB}\)
\(\Leftrightarrow\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{OB}-\overrightarrow{OA}\)
\(\RightarrowĐPCM\)
Câu b ) Bạn làm tương tự câu a , ta có vecto BN = 1/2 (BO +BC ) , rồi là như câu a
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