\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}+\dfrac{2}{5}\overrightarrow{BC}\)

a; Xét ΔCAB có

M,N lần lượt là trung điểm của CB,CA

=>MN là đường trung bình của ΔCAB

=>MN//AB và \(MN=\frac{AB}{2}\)

MN//AB

=>MN//AP và MN//BP

Ta có: \(MN=\frac{AB}{2}\)

\(AP=BP=\frac{AB}{2}\)

Do đó: MN=AP=BP

Xét tứ giác MNAP có

MN//AP

MN=AP

Do đó: MNAP là hình bình hành

=>\(\overrightarrow{MN}=\overrightarrow{PA}\)

mà \(\overrightarrow{BP}=\overrightarrow{PA}\) (P là trung điểm của BA)

nên \(\overrightarrow{MN}=\overrightarrow{BP}\)

Xét ΔBAD có BI là đường trung tuyến

nên \(\overrightarrow{BI}=\dfrac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{BD}\right)\)

=>\(\overrightarrow{BI}=\dfrac{1}{2}\left(\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{BC}\right)\)

\(=\dfrac{1}{2}\left(\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{AC}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{5}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{AC}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{1}{3}\left(5\overrightarrow{BA}+2\overrightarrow{AC}\right)=\dfrac{1}{6}\left(5\overrightarrow{BA}+2\overrightarrow{AC}\right)=\dfrac{5}{6}\left(\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{AC}\right)\)

\(\overrightarrow{BM}=\overrightarrow{BA}+\overrightarrow{AM}\)

\(=\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{AC}\)

=>\(\overrightarrow{BI}=\dfrac{5}{6}\cdot\overrightarrow{BM}\)

=>B,I,M thẳng hàng

Cách 1: Dùng định lý Menelaus đảo:

Từ đề bài, ta có \(\dfrac{BD}{BC}=\dfrac{2}{3}\), \(\dfrac{MC}{MA}=\dfrac{3}{2}\), \(\dfrac{IA}{ID}=1\)

\(\Rightarrow\dfrac{BD}{BC}.\dfrac{MC}{MA}.\dfrac{IA}{ID}=1\)

Theo định lý Menelaus đảo, suy ra B, I, M thẳng hàng.

Cách 2: Dùng vector

 Ta có \(\overrightarrow{BI}=\dfrac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{BD}\right)\)

\(=\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}.\dfrac{2}{3}\overrightarrow{BC}\)

\(=\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{BC}\) 

\(=\dfrac{1}{6}\left(3\overrightarrow{BA}+2\overrightarrow{BC}\right)\)

Lại có \(\overrightarrow{BM}=\dfrac{MC}{AC}\overrightarrow{BA}+\dfrac{MA}{AC}\overrightarrow{BC}\)

\(=\dfrac{3}{5}\overrightarrow{BA}+\dfrac{2}{5}\overrightarrow{BC}\)

\(=\dfrac{1}{5}\left(3\overrightarrow{BA}+2\overrightarrow{BC}\right)\)

\(=\dfrac{6}{5}.\dfrac{1}{6}\left(3\overrightarrow{BA}+2\overrightarrow{BC}\right)\)

\(=\dfrac{6}{5}\overrightarrow{BI}\)

Vậy \(\overrightarrow{BM}=\dfrac{6}{5}\overrightarrow{BI}\), suy ra B, I, M thẳng hàng. 

1: MC=2MB

MC+MB=BC

Do đó: BC=2MB+BM=3BM

=>\(BM=\frac13BC\)

\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)

\(=\overrightarrow{AB}+\frac13\cdot\overrightarrow{BC}=\overrightarrow{AB}+\frac13\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\frac23\cdot\overrightarrow{AB}+\frac13\cdot\overrightarrow{AC}\)

\(=\frac13\left(2\cdot\overrightarrow{AB}+\overrightarrow{AC}\right)\)

2: Xét ΔMAC có MD là đường trung tuyến

nên \(\overrightarrow{MD}=\frac12\left(\overrightarrow{MA}+\overrightarrow{MC}\right)=\frac12\left(\frac23\overrightarrow{AB}+\frac13\cdot\overrightarrow{AC}+\frac23\cdot\overrightarrow{BC}\right)\)

\(=\frac12\left(\frac23\overrightarrow{AC}+\frac13\cdot\overrightarrow{AC}\right)=\frac12\cdot\overrightarrow{AC}=\frac12\left(-\overrightarrow{BA}+\overrightarrow{BC}\right)\)

3: \(\overrightarrow{AE}=\overrightarrow{AB}+\overrightarrow{BE}=\overrightarrow{AB}+\frac12\cdot\overrightarrow{BD}\)

\(=\overrightarrow{AB}+\frac14\cdot\left(\overrightarrow{BA}+\overrightarrow{BC}\right)=\frac34\cdot\overrightarrow{AB}+\frac14\cdot\overrightarrow{BC}\)

\(=\frac34\cdot\overrightarrow{AB}+\frac14\cdot\overrightarrow{BA}+\frac14\cdot\overrightarrow{AC}=\frac12\cdot\overrightarrow{AB}+\frac14\cdot\overrightarrow{AC}\)

\(=\frac14\left(2\cdot\overrightarrow{AB}+\overrightarrow{AC}\right)\)

=>\(\frac{\overrightarrow{AE}}{\overrightarrow{AM}}=\frac14:\frac13=\frac34\)

=>A,E,M thẳng hàng

a: \(\overrightarrow{CN}=\dfrac{1}{2}\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{CB}\)

\(=\dfrac{1}{2}\overrightarrow{CB}+\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{CB}\)

\(=\dfrac{1}{2}\overrightarrow{u}-\overrightarrow{v}\)

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.

\(10< x< 100\Rightarrow10< 3k< 100\)

\(\Rightarrow\dfrac{10}{3}< k< \dfrac{100}{3}\Rightarrow4\le k\le33\)

\(\Rightarrow\sum x=3\left(4+5+...+33\right)=1665\)