Để .... thì pt: \(-x^2+2\left(m+1\right)x-m+1=0\) có 2 nghiệm pb thỏa mãn:
\(\left|x_1-x_2\right|=4\)
\(\Delta'=\left(m+1\right)^2+4\left(-m+1\right)=m^2-2m+5>0;\forall x\)
\(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m-1\end{matrix}\right.\)
Ta có: \(\left(x_1-x_2\right)^2=16\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=16\)
\(\Leftrightarrow4\left(m+1\right)^2-4\left(m-1\right)=16\)
\(\Leftrightarrow m^2+m-2=0\Rightarrow\left[{}\begin{matrix}m=1\\m=-2\end{matrix}\right.\)