\(x^2+\left(2m+1\right)x+m^2+1=0\)
\(\Delta=b^2-4ac=\left(2m-1\right)^2-4\cdot1\cdot\left(m^2+1\right)=4m^2-4m+1-4m^2-4\)
\(\Delta=-4m-3\)
Để phương trình có hai nghiệm \(x_1,x_2\) thì: \(\Delta\ge0\)
\(-4m-3\ge0\Rightarrow-4m\ge3\Rightarrow m\le-\frac{3}{4}\)
Hệ thức Vi-ét:
\(\left\{{}\begin{matrix}x_1+x_2=-\frac{b}{a}=-\frac{2m+1}{1}=-\left(2m+1\right)\\x_1x_2=\frac{c}{a}=\frac{m^2+1}{1}=m^2+1\end{matrix}\right.\)
\(x_1^2+x_2^2=2\left(1+x_1x_2\right)\)
\(x_1^2+2x_1x_2+x_2^2-2x_1x_2-2\left(1+x_1x_2\right)=0\)
\(\left(x_1+x_2\right)^2-2x_1x_2-2-2x_1x_2=0\)
\(\left(x_1+x_2\right)^2-4x_1x_2-2=0\)
\(\left[-\left(2m+1\right)\right]^2-4\left(m^2+1\right)-2=0\)
\(4m^2+4m+1-4m^2-4-2=0\)
\(4m-5=0\)
\(4m=5\)
\(m=\frac{5}{4}\)
Vậy \(m=\frac{5}{4}\)
