\(\left\{{}\begin{matrix}SA\perp\left(ABC\right)\\AB\perp BC\end{matrix}\right.\) \(\Rightarrow BC\perp\left(SAB\right)\)
Kẻ \(AE\perp SB\Rightarrow AE\perp\left(SBC\right)\Rightarrow AE\perp SC\)
Kẻ \(AD\perp SC\Rightarrow SC\perp\left(ADE\right)\)
\(\Rightarrow\widehat{ADE}\) là góc giữa (SAC) và (SBC)
\(\Rightarrow\widehat{ADE}=60^0\)
\(\Rightarrow AE=AD.sin\widehat{ADE}=AD.sin60^0=\dfrac{AD\sqrt{3}}{2}\Rightarrow\dfrac{1}{AE^2}=\dfrac{4}{3AD^2}\)
Áp dụng hệ thức lượng:
\(\dfrac{1}{AE^2}=\dfrac{1}{SA^2}+\dfrac{1}{AB^2}\) ; \(\dfrac{1}{AD^2}=\dfrac{1}{SA^2}+\dfrac{1}{AC^2}\)
\(\Rightarrow\dfrac{1}{SA^2}+\dfrac{1}{AB^2}=\dfrac{4}{3}\left(\dfrac{1}{SA^2}+\dfrac{1}{AC^2}\right)\)
\(\Rightarrow\dfrac{1}{AB^2}=\dfrac{4}{3}.\dfrac{1}{AC^2}+\dfrac{1}{3SA^2}=\dfrac{4}{3}.\dfrac{1}{AB^2+16a^2}+\dfrac{1}{27a^2}\)
Đề có nhầm lẫn đâu không nhỉ, vì phương trình \(\dfrac{1}{x}=\dfrac{4}{3\left(x+16\right)}+\dfrac{1}{27}\) cho nghiệm rất xấu