Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 2
b)\(\overrightarrow{AN}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
\(\overrightarrow{AK}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AN}\right)=\dfrac{1}{2}\left(\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)=\dfrac{3}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\)
d)\(S_{ABC}=24\Leftrightarrow\dfrac{1}{2}AN.BC=24\Leftrightarrow AN=6\left(cm\right)\)
\(\left|\overrightarrow{AB}+\overrightarrow{AC}\right|=\left|2.\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\right|=\left|2\overrightarrow{AN}\right|=2.AN=12\left(cm\right)\)
Bài 3:
b)\(\overrightarrow{BG}=\overrightarrow{BC}+\overrightarrow{CG}=\overrightarrow{BC}+\dfrac{3}{4}\overrightarrow{CA}=\overrightarrow{BC}+\dfrac{3}{4}\left(\overrightarrow{BA}-\overrightarrow{BC}\right)=\dfrac{1}{4}\overrightarrow{BC}+\dfrac{3}{4}\overrightarrow{BA}=\dfrac{1}{4}\overrightarrow{v}+\dfrac{3}{4}\overrightarrow{u}\)
c)Nhìn hình thấy ko thẳng nên đề sai
\(\left\{{}\begin{matrix}\overrightarrow{AM}=\left(x-1;y-3\right)\\\overrightarrow{BM}=\left(x-4;y-2\right)\end{matrix}\right.\)
Tam giác ABM vuông cân tại M khi:
\(\left\{{}\begin{matrix}\overrightarrow{AM}.\overrightarrow{BM}=0\\AM^2=BM^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(x-4\right)+\left(y-3\right)\left(y-2\right)=0\\\left(x-1\right)^2+\left(y-3\right)^2=\left(x-4\right)^2+\left(y-2\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+y^2-5y+10=0\\3x-y=5\end{matrix}\right.\)
Thế \(y=3x-5\) lên pt trên:
\(x^2-5x+\left(3x-5\right)^2-5\left(3x-5\right)+10=0\)
\(\Leftrightarrow x^2-5x+6=0\Rightarrow\left[{}\begin{matrix}x=2\Rightarrow y=1\\x=3\Rightarrow y=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}M\left(2;1\right)\\M\left(3;4\right)\end{matrix}\right.\)
2: ta có: \(\overrightarrow{AB}+\overrightarrow{CD}+\overrightarrow{FE}=\overrightarrow{AE}+\overrightarrow{CB}+\overrightarrow{FD}\)
\(\Leftrightarrow\overrightarrow{AB}+\overrightarrow{FE}+\overrightarrow{EA}=\overrightarrow{CB}+\overrightarrow{FD}+\overrightarrow{DC}\)
\(\Leftrightarrow\overrightarrow{AB}+\overrightarrow{FA}=\overrightarrow{CB}+\overrightarrow{FC}\)
\(\Leftrightarrow\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{FC}-\overrightarrow{FA}\)
\(\Leftrightarrow\overrightarrow{AC}=\overrightarrow{AC}\)(đúng)
