\(3x^2+12x-66=0\)
\(\Leftrightarrow x^2+4x-22=0\)
\(\Leftrightarrow x^2+4x+4-26=0\)
\(\Leftrightarrow\left(x+2\right)^2=26\)
\(\Leftrightarrow x+2=\pm\sqrt{26}\)
\(\Leftrightarrow x=\pm\sqrt{26}-2\)
Áp dụng định lí Py-ta-go ta có :
\(BC^2=AB^2+AC^2\)
mà \(BC^2=\left(\sqrt{3}+1\right)AC^2+\left(\sqrt{3}-1\right)AB.AC\)
\(\rightarrow AB^2+AC^2=\left(\sqrt{3}+1\right)AC^2+\left(\sqrt{3}-1\right)AB.AC\)
\(\rightarrow AB^2-\left(\sqrt{3}-1\right)AB.AC+\left(-\sqrt{3}\right)AC^2=0\)
\(\rightarrow-\sqrt{3}\left(\dfrac{AC}{AB}\right)^2-\left(\sqrt{3}-1\right)\dfrac{AC}{AB}+1=0\)
Do \(a-b+c=-\sqrt{3}+\sqrt{3}-1+1=0\), phương trình có nghiệm
\(\left[{}\begin{matrix}\dfrac{AC}{AB}=-1\left(KTM\right)\\\dfrac{AC}{AB}=-\dfrac{c}{a}=\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)
\(\rightarrow\tan ABC=\dfrac{AC}{AB}=\dfrac{1}{\sqrt{3}}\)
\(\rightarrow ABC=30^o\)
