Để pt có 2 nghiệm pb thì: \(\Delta'>0\Leftrightarrow\left(2m-2\right)^2-3m^2+12m-3>0\)
\(\Leftrightarrow m^2+4m+1>0\)
\(\Leftrightarrow[\begin{matrix}m>-2+\sqrt{3}\\m< -2-\sqrt{3}\end{matrix}\)
theo gt: \(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{2}\)
\(\Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}=\dfrac{x_1+x_2}{2}\)
\(\Rightarrow x_1x_2=2\) (1)
theo viet, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{4}{3}\left(m-1\right)\\x_1x_2=\dfrac{m^2-4m+1}{3}\end{matrix}\right.\) (2)
(1),(2)\(\Rightarrow\dfrac{m^2-4m+1}{3}=2\)
\(\Leftrightarrow m^2-4m+1=6\)
\(\Leftrightarrow m^2-4m-5=0\)
\(\Leftrightarrow\left(m-5\right)\left(m+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=5\\m=-1\end{matrix}\right.\)
kết hợp vs đk\(\Rightarrow m=5\)(t/m)
\(m=-1\)(ko t/m)
Vậy m=5 thì thỏa mãn \(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{1}{2}\left(x_1+x_2\right)\)
