Chương 1: VECTƠ

\(\overrightarrow{BK}=\overrightarrow{BA}+\overrightarrow{AK}=\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{AC}=\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{BC}=\dfrac{2}{3}\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{BC}=\dfrac{1}{3}\left(2\overrightarrow{a}+\overrightarrow{b}\right)\left(1\right)\)\(\overrightarrow{BI}=\overrightarrow{BA}+\overrightarrow{AI}=\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{AM}=\overrightarrow{BA}+\dfrac{1}{2}.\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\overrightarrow{BA}+\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}=\dfrac{3}{4}\overrightarrow{BA}+\dfrac{1}{4}\left(\overrightarrow{AB}+\overrightarrow{BC}\right)=\dfrac{3}{4}\overrightarrow{BA}+\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{BC}=\dfrac{2}{4}\overrightarrow{BA}+\dfrac{1}{4}\overrightarrow{BC}=\dfrac{1}{4}\left(2\overrightarrow{a}+\overrightarrow{b}\right)\left(2\right)\)từ (1) và (2) -> \(\overrightarrow{BK}và\overrightarrow{BI}\) cùng phương -> B,K,I thẳng hàng

Chọn A

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Bài 1:

Ta có:

\(\overrightarrow{AB}-\overrightarrow{CB}+\overrightarrow{CD}-\overrightarrow{ED}\)

\(=(\overrightarrow{AC}+\overrightarrow{CB})-\overrightarrow{CB}+(\overrightarrow{CE}+\overrightarrow{ED})-\overrightarrow{ED}\)

\(=\overrightarrow{AC}+\overrightarrow{CE}=\overrightarrow{AE}\)

Bài 2: Đề bài không rõ ràng, bạn xem lại hộ mình nhé.

gọi D(x;y)

A là trọng tâm của tam giác BCD ⇔\(\left\{{}\begin{matrix}-1=\dfrac{1+1+x}{3}\\1=\dfrac{3+\left(-1\right)+y}{3}\end{matrix}\right.\) ⇔\(\left\{{}\begin{matrix}x=-5\\y=1\end{matrix}\right.\) vậy D(-5;1)

\(\cos BAC=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)

\(\Leftrightarrow\dfrac{4^2+AC^2-7^2}{2\cdot4\cdot AC}=\dfrac{2}{3}\)

\(\Leftrightarrow3AC^2-99=8AC\)

\(\Leftrightarrow3AC^2-8AC-99=0\)

\(\text{Δ}=\left(-8\right)^2-4\cdot3\cdot\left(-99\right)=1252>0\)

Do đó: \(\left[{}\begin{matrix}AC=\dfrac{4+\sqrt{313}}{3}\left(nhận\right)\\AC=\dfrac{4-\sqrt{313}}{3}\left(loại\right)\end{matrix}\right.\)