Question

tìm tất cả các giá trị của m để phương trình

\(mx^2-2\left(m-1\right)x+3\left(m-2\right)=0\) có 2 nghiệm \(x_1,x_2\) thỏa mãn \(2x_1+x_2=1\)

Answer 1

\(mx^2-2\left(m-1\right)x+3\left(m-2\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}m\ne0\\\Delta'=\left(m-1\right)^2-3m\left(m-2\right)\ge0\end{matrix}\right.\) \(\Rightarrow\frac{2-\sqrt{6}}{2}\le m\le\frac{2+\sqrt{6}}{2};m\ne0\)

Khi đó ta có: \(\left\{{}\begin{matrix}x_1+x_2=\frac{2\left(m-1\right)}{m}\\x_1x_2=\frac{3\left(m-2\right)}{m}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=\frac{2\left(m-1\right)}{m}\\2x_1+x_2=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=\frac{2-m}{m}\\x_2=\frac{3m-4}{m}\end{matrix}\right.\)

\(\Rightarrow\left(\frac{2-m}{m}\right)\left(\frac{3m-4}{m}\right)=\frac{3\left(m-2\right)}{m}\Leftrightarrow3\left(m-2\right)+\left(m-2\right)\left(3m-4\right)=0\)

\(\Leftrightarrow\left(m-2\right)\left(3m-1\right)=0\) \(\Rightarrow\left[{}\begin{matrix}m=2\\m=\frac{1}{3}\end{matrix}\right.\)