\(x^2-\left(m-4\right)x+m+4=0\)
\(\text{Δ}=\left[-\left(m-4\right)\right]^2-4\cdot1\cdot\left(m+4\right)\)
\(=\left(m-4\right)^2-4\left(m+4\right)\)
\(=m^2-8m+16-4m-16=m^2-12m=m\left(m-12\right)\)
Để phương trình có hai nghiệm thì Δ>=0
=>m(m-12)>=0
=>\(\left[{}\begin{matrix}m>=12\\m< =0\end{matrix}\right.\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left[-\left(m-4\right)\right]}{1}=m-4\\x_1\cdot x_2=\dfrac{c}{a}=m+4\end{matrix}\right.\)
\(x_1\left(x_1-1\right)+x_2\left(x_2-1\right)=18\)
\(\Leftrightarrow\left(x_1^2+x_2^2\right)-\left(x_1+x_2\right)=18\)
=>\(\left(x_1+x_2\right)^2-2x_1x_2-\left(x_1+x_2\right)=18\)
=>\(\left(m-4\right)^2-2\left(m+4\right)-\left(m-4\right)=18\)
=>\(m^2-8m+16-2m-8-m+4=18\)
=>\(m^2-11m+12-18=0\)
=>\(m^2-11m-6=0\)
=>\(\left[{}\begin{matrix}m=\dfrac{11+\sqrt{145}}{2}\left(loại\right)\\m=\dfrac{11-\sqrt{145}}{2}\left(nhận\right)\end{matrix}\right.\)