Question

giúp mình hình 4 với ạ!!!!!!

Answer 1

Hình 4:

Áp dụng HTL:

\(AH^2=BH\cdot HC\Rightarrow BH\cdot HC=6^2=36\Rightarrow xy=36\)

Mà \(BH+HC=BC\Rightarrow x+y=25\)

Ta có hệ \(\left\{{}\begin{matrix}xy=36\\x+y=25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{36}{y}\\\dfrac{36}{y}+y=25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{36}{y}\\y^2-25y+36=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{36}{y}\\\Delta=625-144=481\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{36}{y}\\\left[{}\begin{matrix}y=\dfrac{25-\sqrt{481}}{2}\\y=\dfrac{25+\sqrt{481}}{2}\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow...\)