\(\left(mx-2\right)\left(2mx-x+1\right)=0\)
\(\Leftrightarrow\left(mx-2\right)\left[\left(2m-1\right)x+1\right]=0\)
\(\Leftrightarrow\left(2m-1\right).m.x^2+mx-2x\left(2m-1\right)-2=0\)
\(\Leftrightarrow\left(2m^2-m\right)x^2+x\left[m-2\left(2m-1\right)\right]-2=0\)
\(\Leftrightarrow\left(2m^2-m\right)x^2+\left(2-3m\right)x-2=0\)
\(\Delta=\left(2-3m\right)^2-4\left(2m^2-m\right).\left(-2\right)\)
\(\Delta=4-12m+9m^2+16m^2-8m\)
\(\Delta=25m^2-20m+4=\left(5m-2\right)^2\)
Để pt có nghiệm kép khi \(\left\{{}\begin{matrix}a\ne0\\\Delta=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2m^2-m\ne0\\\left(5m-2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ne0;\dfrac{1}{2}\\m=\dfrac{2}{5}\end{matrix}\right.\)
vậy \(m=\dfrac{2}{5}\) thì pt có nghiệm kép