\(\Delta=4\left(m+1\right)^2-4\left(m^2+m+1\right)\)\(=4m\)
Để pt có 2 nghiệm phân biệt thì \(\Delta>0\Leftrightarrow4m>0\Leftrightarrow m>0\)
Theo Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1.x_2=m^2+m+1\end{matrix}\right.\)
Theo bài ra: \(x_1^2+x_2^2=3x_1.x_2-1\)\(\Leftrightarrow\left(x_1+x_2\right)^2-5x_1.x_2+1=0\)
Kết hợp Vi-ét: \(4\left(m+1\right)^2-5\left(m^2+m+1\right)+1=0\)
\(\Leftrightarrow m^2-3m=0\)\(\Rightarrow\left[{}\begin{matrix}m=0\left(l\right)\\m=3\left(tm\right)\end{matrix}\right.\)
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