\(\left(m-2\right)x^2-2x+1-2m=0\)
Ta có: △ = \(b^2-4ac\)
= \(\left(-2\right)^2-4.\left(m-2\right).\left(1-2m\right)\)
= 4 - 4.( \(-2m^2+5m-2\) )
= \(8m^2-20m+12\)
+ Nếu △ > 0
⇔ \(8m^2-20m+12>0\)
⇔ \(\left[{}\begin{matrix}m< 1\\m>\dfrac{3}{2}\end{matrix}\right.\)
Phương trình có hai nghiệm phân biệt
\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}\)
= \(\dfrac{2+\sqrt{8m^2-20m+12}}{2\left(m-2\right)}\)
= \(\dfrac{1+\sqrt{2m^2-5m+3}}{m-2}\)
\(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}\)
= \(\dfrac{2-\sqrt{8m^2-20m+12}}{2\left(m-2\right)}\)
= \(\dfrac{1-\sqrt{2m^2-5m+3}}{m-2}\)
+ Nếu Δ = 0
⇔ \(8m^2-20m+12=0\)
⇔ \(\left[{}\begin{matrix}m=\dfrac{3}{2}\\m=1\end{matrix}\right.\)
Phương trình có nghiệm kép
\(x_1=x_2=\dfrac{-b}{2a}=\dfrac{2}{2\left(m-2\right)}=\dfrac{1}{m-2}\)
+ Nếu Δ < 0
⇔ \(8m^2-20m+12< 0\)
⇔ 1<m <\(\dfrac{3}{2}\)