Tự xử lí delta nha
Theo vi-et: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1.x_2=-\left(m-1\right)\left(m-3\right)\end{matrix}\right.\)
Theo đề: \(\frac{1}{4}.\left(2m\right)^2-\left(m-1\right)\left(m-3\right)-2.2m+3=0\)
<=> \(m^2-m^2+4m-3-4m+3=0\) (TM)
Vậy vs mọi m thỏa delta thì ...