\(\Delta=\left(-2m\right)^2-4\cdot1\cdot4=4m^2-16\)
Để phương trình có hai nghiệm thì \(\Delta>=0\)
=>\(4m^2-16>=0\)
=>\(m^2>=4\)
=>\(\left[{}\begin{matrix}m>=2\\m< =-2\end{matrix}\right.\)
Theo Vi-et, ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m\\x_1x_2=\dfrac{c}{a}=4\end{matrix}\right.\)
\(\left(x_1+1\right)^2+\left(x_2+1\right)^2=2\)
=>\(x_1^2+x_2^2+2\left(x_1+x_2\right)=0\)
=>\(\left(x_1+x_2\right)^2-2x_1x_2+2\left(x_1+x_2\right)=0\)
=>\(\left(2m\right)^2-2\cdot4+2\cdot2m=0\)
=>\(4m^2+4m-8=0\)
=>\(m^2+m-2=0\)
=>(m+2)(m-1)=0
=>\(\left[{}\begin{matrix}m=-2\left(nhận\right)\\m=1\left(loại\right)\end{matrix}\right.\)
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