Toán 10

N

nhuquynh_377

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N

noinhobinhyen

Đặt AB⃗=x⃗;AC⃗=y⃗⇒∣x⃗∣=∣y⃗∣=a\vec{AB}=\vec{x} ; \vec{AC}=\vec{y} \Rightarrow |\vec{x}|=|\vec{y}|=a

x⃗y⃗=−a22\vec{x}\vec{y}=\dfrac{-a^2}{2}

BC⃗=y⃗−x⃗\vec{BC}=\vec{y}-\vec{x}

Ta có

$T=\vec{AB}.\vec{CB}+\vec{CB}.\vec{CA} +\vec{AC}.\vec{BA}=\vec{x}(\vec{x}-\vec{y})+
(\vec{x}-\vec{y}).(-\vec{y})+\vec{y}.(-\vec{x})$

....

Quỳnh làm tiếp nha !!
 
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