\(x^2-4mx+m^2-2m+1=0\)
\(\Delta'=\left(-2m\right)^2-\left(m^2-2m+1\right)=4m^2-m^2+2m-1=3m^2+2m-1\)
* Pt có 2 nghiệm phân biệt \(\Leftrightarrow\Delta'>0\Leftrightarrow3m^2+2m-1>0\Leftrightarrow\left[{}\begin{matrix}m< -1\\m>\frac{1}{3}\end{matrix}\right.\)
* Ta có: \(a.c=m^2-2m+1=\left(m-1\right)^2\ge0\)
=> x1, x2 không thể có hai nghiệm trái dấu
* Pt có 2 nghiệm phân biệt khi \(\left[{}\begin{matrix}m< -1\\m>\frac{1}{3}\end{matrix}\right.\) (1)
Theo định lí Vi-ét: \(\left\{{}\begin{matrix}S=x_1+x_2=4m\\P=x_1x_2=\left(m-1\right)^2\end{matrix}\right.\)
\(\left|\sqrt{x_1}-\sqrt{x_2}\right|=1\)
\(\Rightarrow\left(\sqrt{x_1}-\sqrt{x_2}\right)^2=1\)
\(\Leftrightarrow x_1-2\sqrt{x_1x_2}+x_2=1\)
\(\Leftrightarrow x_1+x_2-2\sqrt{x_1x_2}=1\)
\(\Leftrightarrow4m-2\sqrt{\left(m-1\right)^2}=1\)
\(\Leftrightarrow-2\left|m-1\right|=1-4m\)
\(\Leftrightarrow\left|m-1\right|=2m-\frac{1}{2}\)
ĐK: \(2m-\frac{1}{2}\ge0\Leftrightarrow m\ge\frac{1}{4}\)
+ TH1: \(m-1=2m-\frac{1}{2}\Leftrightarrow m=-\frac{1}{2}\) (0 t/m)
+ TH2: \(m-1=\frac{1}{2}-2m\Leftrightarrow3m=\frac{3}{2}\Leftrightarrow m=\frac{1}{2}\) (t/m) (2)
(1),(2): Vậy \(m=\frac{1}{2}\) thỏa mãn đề bài
