\(x^2-\left(2m-1\right)x+2m-2=0\left(1\right)\)
a) Để \(\left(1\right)\) có \(2\) nghiệm phân biệt thỏa \(x_1^2+x_2^2=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(2m-1\right)^2-8m+8>0\\x_1^2+x_2^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4m^2-4m+1-8m+8>0\\\left(x_1+x_2\right)^2-2x_1x_2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4m^2-12m+9>0\\4m^2-4m+1-4m+4=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2m-3\right)^2>0\\4m^2-8m+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ne\dfrac{3}{2}\left(1\right)\\m=1\end{matrix}\right.\) \(\Leftrightarrow m=1\)
b) Để \(\left(1\right)\) có \(2\) nghiệm phân biệt thỏa \(x_1< \dfrac{3}{2}< x_2\)
\(\Leftrightarrow\left(x_1-\dfrac{3}{2}\right)\left(x_2-\dfrac{3}{2}\right)< 0\)
\(\Leftrightarrow x_1x_2-\dfrac{3}{2}\left(x_1+x_2\right)+\dfrac{9}{4}< 0\)
\(\Leftrightarrow2m-2-\dfrac{3}{2}.\left(2m-1\right)+\dfrac{9}{4}< 0\)
\(\Leftrightarrow-m-2+\dfrac{3}{2}+\dfrac{9}{4}< 0\)
\(\Leftrightarrow m>\dfrac{7}{4}\) thỏa \(\left(1\right)\)
Vậy \(m>\dfrac{7}{4}\) thỏa yêu cầu đề bài
