Question

Tìm m để phương trình có hai nghiệm phân biệt?
a,(m-1)\(x^2\) -2m+1=0
b,\(x^2-\left(2m+1\right)x+2m=0\)
c,\(x^2-mx+m-1=0\)

Answer 1

a/ \(\left\{{}\begin{matrix}m\ne1\\\Delta'=0-\left(m-1\right)\left(-2m+1\right)>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ne1\\\left(m-1\right)\left(2m-1\right)>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ne1\\\left[{}\begin{matrix}m>1\\m< \frac{1}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m>1\\m< \frac{1}{2}\end{matrix}\right.\)

b/ \(\Delta=\left(2m+1\right)^2-8m>0\)

\(\Leftrightarrow\left(2m-1\right)^2>0\Rightarrow m\ne\frac{1}{2}\)

c/ \(\Delta=m^2-4\left(m-1\right)=\left(m-2\right)^2>0\Rightarrow m\ne2\)