\(\Delta=\left(2m-1\right)^2-4\left(3m-4\right)=4m^2-16m+17=4\left(m-2\right)^2+1>0;\forall m\)
\(\Rightarrow\) Pt đã cho luôn có 2 nghiệm pb với mọi m
b.
Để biểu thức đề bài xác định \(\Rightarrow x_1;x_2\ne1\Leftrightarrow1+2m-1+3m-4\ne0\Rightarrow m\ne\dfrac{4}{5}\)
Khi đó theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-2m+1\\x_1x_2=3m-4\end{matrix}\right.\)
\(\dfrac{x_1^2}{1-x_1}+\dfrac{x_2^2}{1-x_2}=2\Leftrightarrow\dfrac{x_1^2-x_1^2x_2+x_2^2-x_1x_2^2}{\left(1-x_1\right)\left(1-x_2\right)}=2\)
\(\Leftrightarrow\dfrac{\left(x_1+x_2\right)^2-2x_1x_2-x_1x_2\left(x_1+x_2\right)}{x_1x_2-\left(x_1+x_2\right)+1}=2\)
\(\Leftrightarrow\dfrac{\left(-2m+1\right)^2-2\left(3m-4\right)-\left(3m-4\right)\left(-2m+1\right)}{3m-4-\left(-2m+1\right)+1}=2\)
\(\Leftrightarrow\dfrac{10m^2-21m+13}{5m-4}=2\)
\(\Rightarrow10m^2-21m+13=10m-8\)
\(\Leftrightarrow10m^2-31m+21=0\Rightarrow\left[{}\begin{matrix}m=1\\m=\dfrac{21}{10}\end{matrix}\right.\)
