\(\text{Δ}=\left[4\left(m-1\right)\right]^2-4\cdot3\cdot\left(m^2-4m+1\right)\)
\(=\left(4m-4\right)^2-12\left(m^2-4m+1\right)\)
\(=16m^2-32m+16-12m^2+48m-12\)
\(=4m^2+16m+4\)
Để phương trình có hai nghiệm thì Δ>=0
=>\(4m^2+16m+4>=0\)
=>\(m^2+4m+1>=0\)
=>\(\left(m+2\right)^2>=3\)
=>\(\left[{}\begin{matrix}m+2>=\sqrt{3}\\m+2< =-\sqrt{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}m>=\sqrt{3}-2\\m< =-\sqrt{3}-2\end{matrix}\right.\)
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-4\left(m-1\right)}{3}\\x_1x_2=\dfrac{c}{a}=\dfrac{m^2-4m+1}{3}\end{matrix}\right.\)
\(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{1}{2}\left(x_1+x_2\right)\)
=>\(\dfrac{x_1+x_2}{x_1x_2}=\dfrac{1}{2}\left(x_1+x_2\right)\)
=>\(\dfrac{-4\left(m-1\right)}{3}:\dfrac{m^2-4m+1}{3}=\dfrac{1}{2}\cdot\dfrac{-4\left(m-1\right)}{3}\)
=>\(\dfrac{-4\left(m-1\right)}{m^2-4m+1}=\dfrac{-2\left(m-1\right)}{3}\)
=>\(\dfrac{2\left(m-1\right)}{m^2-4m+1}=\dfrac{m-1}{3}\)
=>\(\left(m-1\right)\left(\dfrac{2}{m^2-4m+1}-\dfrac{1}{3}\right)=0\)
=>\(\left[{}\begin{matrix}m-1=0\\\dfrac{2}{m^2-4m+1}-\dfrac{1}{3}=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}m-1=0\\\dfrac{2}{m^2-4m+1}=\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}m=1\\m^2-4m+1=6\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}m=1\\m^2-4m-5=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}m=1\\\left(m-5\right)\left(m+1\right)=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}m=1\left(nhận\right)\\m=5\left(nhận\right)\\m=-1\left(loại\right)\end{matrix}\right.\)
