Có vẻ không đúng.

Giả sử \(\overrightarrow{AB}+\overrightarrow{MB}+\overrightarrow{MA}=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{MB}+\left(\overrightarrow{MA}+\overrightarrow{AB}\right)=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{MB}+\overrightarrow{MB}=\overrightarrow{0}\)

\(\Leftrightarrow2\overrightarrow{MB}=\overrightarrow{0}\)

\(\Leftrightarrow M\equiv B\) (Vô lí)

Hình vẽ:

a, Chứng minh \(\overrightarrow{AB}+\overrightarrow{BM}+\overrightarrow{MA}=\overrightarrow{0}\)

Ta có \(\overrightarrow{AB}+\overrightarrow{BM}+\overrightarrow{MA}=\overrightarrow{BM}+\left(\overrightarrow{MA}+\overrightarrow{AB}\right)=\overrightarrow{BM}+\overrightarrow{MB}=\overrightarrow{0}\)

b, Gọi H là trung điểm \(MC\)

Ta có \(AM=\sqrt{AC^2-MC^2}=\sqrt{4a^2-a^2}=a\sqrt{3}\)

\(AH=\sqrt{AM^2+MH^2}=\sqrt{\left(a\sqrt{3}\right)^2+\left(\dfrac{a}{2}\right)^2}=a.\dfrac{\sqrt{13}}{2}\)

\(\left|\overrightarrow{AM}+\overrightarrow{AC}\right|=\left|2\overrightarrow{AH}\right|=2AH=a\sqrt{13}\)

c, Gọi D là trung điểm AB

\(3\overrightarrow{NA}+3\overrightarrow{NB}+2\overrightarrow{NC}=3\left(\overrightarrow{NA}+\overrightarrow{NB}\right)+2\overrightarrow{NC}=6\overrightarrow{ND}+2\overrightarrow{NC}=\overrightarrow{0}\)

\(\Rightarrow\overrightarrow{NC}=3\overrightarrow{DN}\)

Vậy N thuộc đoạn CD sao cho \(CN=\dfrac{3}{4}CD\)

Tham khảo:

a)  \(\)\(\overrightarrow {BA}  + \overrightarrow {AC}  = \overrightarrow {BC}  \Rightarrow \left| {\overrightarrow {BC} } \right| = BC = a\)

b) Dựng hình bình hành ABDC, giao điểm của hai đường chéo là O ta có:

\(\overrightarrow {AB}  + \overrightarrow {AC}  = \overrightarrow {AD} \)

\(AD = 2AO = 2\sqrt {A{B^2} - B{O^2}}  = 2\sqrt {{a^2} - {{\left( {\frac{a}{2}} \right)}^2}}  = a\sqrt 3 \)

\( \Rightarrow \left| {\overrightarrow {AB}  + \overrightarrow {AC} } \right| = \left| {\overrightarrow {AD} } \right| = AD = a\sqrt 3 \)

c) \(\overrightarrow {BA}  - \overrightarrow {BC}  = \overrightarrow {BA}  + \overrightarrow {CB}  = \overrightarrow {CB}  + \overrightarrow {BA}  = \overrightarrow {CA} \)

\( \Rightarrow \left| {\overrightarrow {BA}  - \overrightarrow {BC} } \right| = \left| {\overrightarrow {CA} } \right| = CA = a\)

a.

\(P=cos120^0+cos120^0+cos120^0=-\dfrac{3}{2}\)

b.

\(A=\dfrac{\dfrac{sinx}{cosx}-\dfrac{cosx}{cosx}}{\dfrac{sinx}{cosx}+\dfrac{cosx}{cosx}}=\dfrac{tanx-1}{tanx+1}=\dfrac{2-1}{2+1}=\dfrac{1}{3}\)

c.

\(A=\dfrac{cos\left(720+30\right)+sin\left(360+60\right)}{sin\left(-360+30\right)-cos\left(-360-30\right)}=\dfrac{cos30+sin60}{sin30-cos30}=-3-\sqrt{3}\)

a) \(\overrightarrow{AB}.\overrightarrow{AC}=0\) do \(AB\perp AC\).
b)
\(BC=\sqrt{AB^2+AC^2}=\sqrt{a^2+a^2}=\sqrt{2}a\).
\(\overrightarrow{BA}.\overrightarrow{BC}=BA.BC.cos\left(\overrightarrow{BA},\overrightarrow{BC}\right)=a.\sqrt{2}a.cos45^o=a^2\).
c) \(\overrightarrow{AB}.\overrightarrow{BC}=-\overrightarrow{BA}.\overrightarrow{BC}=-a^2\).

a) Có
\(\overrightarrow{BC}^2=\left(\overrightarrow{BA}+\overrightarrow{AC}\right)^2=\overrightarrow{BA}^2+\overrightarrow{AC}^2+2\overrightarrow{BA}.\overrightarrow{AC}\)
\(=\overrightarrow{BA}^2+\overrightarrow{AC}^2-2\overrightarrow{AB}.\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{AB}.\overrightarrow{AC}=\dfrac{\overrightarrow{BA}^2+\overrightarrow{AC}^2-\overrightarrow{BC^2}}{2}=\dfrac{5^2+8^2-7^2}{2}=20\).
\(cos\widehat{BAC}=\dfrac{\overrightarrow{AB}.\overrightarrow{AC}}{\left|\overrightarrow{AB}\right|.\left|\overrightarrow{AC}\right|}=\dfrac{20}{5.8}=\dfrac{1}{2}\).
Vì vậy \(\widehat{BAC}=60^o\).
b) Tương tự:
\(\overrightarrow{CA}.\overrightarrow{CB}=\dfrac{CA^2+CB^2-AB^2}{2}=\dfrac{7^2+8^2-5^2}{2}=44\).