kẻ AC cắt MN tại I
ta có \(\dfrac{AM}{MD}=\dfrac{BN}{NC}\left(=\dfrac{1}{2}\right)\Rightarrow\)MN//DC//AB(định lí Talet đảo)
ta có:
\(\dfrac{AM}{MD}=\dfrac{1}{2}\Rightarrow\dfrac{AM}{AD}=\dfrac{AM}{AM+MD}=\dfrac{1}{1+2}=\dfrac{1}{3}\\ \Rightarrow\dfrac{BN}{BC}=\dfrac{1}{3}\Rightarrow\dfrac{NC}{BC}=\dfrac{2}{3}\)
tam giác ADC có MI//DC nên :\(\dfrac{MI}{DC}=\dfrac{AM}{AD}\Rightarrow\dfrac{MI}{\left(x+2\right)\left(x-2\right)}=\dfrac{1}{3}\Rightarrow\left(x+2\right)\left(x-2\right)=3MI\)
tam giac CAB có IN//AB nên:
\(\dfrac{NI}{AB}=\dfrac{BN}{BC}\Rightarrow\dfrac{NI}{x+4}=\dfrac{2}{3}\Rightarrow2\left(x+4\right)=3NI\)
\(\left(x+2\right)\left(x-2\right)+2\left(x+4\right)=3MI+3NI\\ \Leftrightarrow x^2-4+2x+8=3\left(MI+NI\right)=3MN=39\\ \Leftrightarrow x^2+2x+4-39=0\\ \Leftrightarrow x^2+2x-35=0\\ \Leftrightarrow\left(x-5\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)
Vậy x=5 hoặc x=-7
