+) Ta có: \(AB \bot AC \Rightarrow \overrightarrow {AB}  \bot \overrightarrow {AC}  \Rightarrow \overrightarrow {AB} .\overrightarrow {AC}  = 0\)

+) \(\overrightarrow {AC} .\overrightarrow {BC}  = \left| {\overrightarrow {AC} } \right|.\left| {\overline {BC} } \right|.\cos \left( {\overrightarrow {AC} ,\overrightarrow {BC} } \right)\)

Ta có: \(BC = \sqrt {A{B^2} + A{C^2}}  = \sqrt 2  \Leftrightarrow \sqrt {2A{C^2}}  = \sqrt 2 \)\( \Rightarrow AC = 1\)

\( \Rightarrow \overrightarrow {AC} .\overrightarrow {BC}  = 1.\sqrt 2 .\cos \left( {45^\circ } \right) = 1\)

+) \(\overrightarrow {BA} .\overrightarrow {BC}  = \left| {\overrightarrow {BA} } \right|.\left| {\overrightarrow {BC} } \right|.\cos \left( {\overrightarrow {BA} ,\overrightarrow {BC} } \right) = 1.\sqrt 2 .\cos \left( {45^\circ } \right) = 1\)

\(T=\overrightarrow{GA}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)+\overrightarrow{GB}.\overrightarrow{CA}+\overrightarrow{GC}.\overrightarrow{AB}\)

\(=\overrightarrow{AB}\left(\overrightarrow{GC}-\overrightarrow{GA}\right)+\overrightarrow{AC}\left(\overrightarrow{GA}-\overrightarrow{GB}\right)\)

\(=\overrightarrow{AB}\left(\overrightarrow{GC}+\overrightarrow{AG}\right)+\overrightarrow{AC}\left(\overrightarrow{GA}+\overrightarrow{BG}\right)\)

\(=\overrightarrow{AB}.\overrightarrow{AC}+\overrightarrow{AC}.\overrightarrow{BA}\)

\(=0\)

Ta có: CB= a√2; = 45⁰

Vậy = -. = -||: ||. cos45⁰ = -a.a√2.

=> = -a²

D là điểm nào bạn?

1.

\(\Leftrightarrow x^2-3x+1+\dfrac{\sqrt{3}}{3}\sqrt{\left(x^2+x+1\right)\left(x^2-x+1\right)}=0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+1}=a>0\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\)

\(\Rightarrow2b^2-a^2+\dfrac{\sqrt{3}}{3}ab=0\)

\(\Leftrightarrow\left(\sqrt{3}b-a\right)\left(2b+\sqrt{3}a\right)=0\)

\(\Leftrightarrow a=\sqrt{3}b\)

\(\Leftrightarrow\sqrt{x^2+x+1}=\sqrt{3}.\sqrt{x^2-x+1}\)

\(\Leftrightarrow x^2+x+1=3x^2-3x+3\)

\(\Leftrightarrow2x^2-4x+2=0\)

\(\Leftrightarrow x=1\)

Bài 2:

Đặt \(AB=x>0\) 

\(AG=\dfrac{1}{2}BC=\dfrac{1}{2}\sqrt{a^2+x^2}\)

\(CG=\dfrac{2}{3}\sqrt{\left(\dfrac{AB}{2}\right)^2+AC^2}=\dfrac{2}{3}\sqrt{\dfrac{x^2}{4}+a^2}\)

\(BG=\dfrac{2}{3}\sqrt{AB^2+\left(\dfrac{AC}{2}\right)^2}=\dfrac{2}{3}\sqrt{x^2+\dfrac{a^2}{4}}\)

Ta có:

\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\Leftrightarrow\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{AG}\)

\(\Leftrightarrow GB^2+GC^2+2GB.GC.cos\left(\overrightarrow{GB};\overrightarrow{GC}\right)=AG^2\)

\(\Leftrightarrow cos\left(\overrightarrow{GB};\overrightarrow{GC}\right)=\dfrac{AG^2-BG^2-CG^2}{2GB.GC}\)

\(=\dfrac{\dfrac{a^2+x^2}{4}-\left[\dfrac{4}{9}\left(\dfrac{x^2}{4}+a^2\right)+\dfrac{4}{9}\left(\dfrac{a^2}{4}+x^2\right)\right]}{\dfrac{2}{9}\sqrt{\left(a^2+4x^2\right)\left(x^2+4a^2\right)}}\)

\(=-\dfrac{11}{4}.\dfrac{x^2+a^2}{2\sqrt{\left(a^2+4x^2\right)\left(x^2+4a^2\right)}}\le-\dfrac{11}{4}.\dfrac{x^2+a^2}{5\left(x^2+a^2\right)}=-\dfrac{11}{20}\)

Dấu "=" xảy ra khi \(a=x\Leftrightarrow AB=a\)

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

b: lấy điểm H sao cho \(\overrightarrow{AH}=\overrightarrow{GC}\)

\(\overrightarrow{AH}=\overrightarrow{GC}\)

=>AH//GC và AH=GC

Xét tứ giác AHCG có

AH//CG

AH=GC

Do đó: AHCG là hình bình hành

ΔABC đều có G là trọng tâm

nên \(AG=GB=GC=\dfrac{a\sqrt{3}}{3}\)

\(\left|\overrightarrow{AB}-\overrightarrow{GC}\right|=\left|\overrightarrow{AB}-\overrightarrow{AH}\right|\)

\(=\left|\overrightarrow{HA}+\overrightarrow{AB}\right|=\left|\overrightarrow{HB}\right|=HB\)

AHCG là hình bình hành

=>HC=AG và HC//AG

=>\(HC=\dfrac{a\sqrt{3}}{3}\)

ΔABC đều có G là trọng tâm

nên GB=GC=GA

GB=GC

AB=AC

Do đó: AG là đường trung trực của BC

=>AG\(\perp\)BC

mà CH//AG

nên CH\(\perp\)CB

=>ΔCHB vuông tại C

=>\(BH^2=HC^2+BC^2\)

=>\(BH^2=\left(\dfrac{a\sqrt{3}}{3}\right)^2+a^2=a^2+\dfrac{1}{3}a^2=\dfrac{4}{3}a^2\)

=>\(BH=a\cdot\dfrac{2\sqrt{3}}{3}\)

=>\(\left|\overrightarrow{AB}-\overrightarrow{GC}\right|=BH=\dfrac{2a\sqrt{3}}{3}\)

Bài 3: 

Tham khảo:

Ta có: \(AC = BD = \sqrt {A{B^2} + B{C^2}}  = \sqrt {{a^2} + {a^2}}  = a\sqrt 2 \)

+) \(AB \bot AD \Rightarrow \overrightarrow {AB}  \bot \overrightarrow {AD}  \Rightarrow \overrightarrow {AB} .\overrightarrow {AD}  = 0\)

+) \(\overrightarrow {AB} .\overrightarrow {AC}  = \left| {\overrightarrow {AB} } \right|.\left| {\overrightarrow {AC} } \right|.\cos \left( {\overrightarrow {AB} ,\overrightarrow {AC} } \right) = a.a\sqrt 2.\cos 45^\circ  = a^2\)

+) \(\overrightarrow {AC} .\overrightarrow {CB}  = \left| {\overrightarrow {AC} } \right|.\left| {\overrightarrow {CB} } \right|.\cos \left( {\overrightarrow {AC} ,\overrightarrow {CB} } \right) = a\sqrt 2 .a.\cos 135^\circ  =  - {a^2}\)

+) \(AC \bot BD \Rightarrow \overrightarrow {AC}  \bot \overrightarrow {BD}  \Rightarrow \overrightarrow {AC} .\overrightarrow {BD}  = 0\)

Chú ý

\(\overrightarrow {a}  \bot \overrightarrow {b}  \Leftrightarrow \overrightarrow {a} .\overrightarrow {b}  = 0\)

\(a,\) \(\overrightarrow{IA}=2\overrightarrow{IB}-4\overrightarrow{IC}\)

\(\overrightarrow{IA}=2\overrightarrow{IB}-2\overrightarrow{IC}-2\overrightarrow{IC}=2\overrightarrow{CB}-2\overrightarrow{IC}\)

\(=2\left(\overrightarrow{AB}-\overrightarrow{AC}\right)-2\left(\overrightarrow{AC}-\overrightarrow{AI}\right)\)

\(\overrightarrow{IA}=2\overrightarrow{AB}-2\overrightarrow{AC}-2\overrightarrow{AC}+2\overrightarrow{AI}\)

\(\overrightarrow{IA}=\dfrac{2}{3}\overrightarrow{AB}-\dfrac{4}{3}\overrightarrow{AC}\)

\(b,\overrightarrow{IJ}=\overrightarrow{AJ}-\overrightarrow{AI}=\dfrac{2}{3}\overrightarrow{AB}+\overrightarrow{IA}=\dfrac{2}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AB}-\dfrac{4}{3}\overrightarrow{AC}=\dfrac{4}{3}\left(\overrightarrow{AB}-\overrightarrow{AC}\right)\left(1\right)\)

\(\overrightarrow{JG}=\overrightarrow{AG}-\overrightarrow{AJ}=\dfrac{2}{3}\overrightarrow{AM}-\dfrac{2}{3}\overrightarrow{AB}\)\((\) \(\) \(M\)  \(trung\) \(điểm\) \(BC)\)

\(\overrightarrow{JG}=\dfrac{\overrightarrow{AB}+\overrightarrow{AC}}{3}-\dfrac{2}{3}\overrightarrow{AB}=-\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}=-\dfrac{1}{3}\left(\overrightarrow{AB}-\overrightarrow{AC}\right)\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow\overrightarrow{IJ}=-4\overrightarrow{JG}\Rightarrow I,J,G\) \(thẳng\) \(hàng\)