3.b)
\(\Delta=b^2-4ac=\left[-\left(m+5\right)\right]^2-4.2.\left(-3m^2+10m-3\right)\)
\(=m^2+10m+25+24m^2-80m+24\)
\(=25m^2-70m+49\)
\(=\left(5m-7\right)^2\ge0\forall m\in R\)
Theo định lý Vi-ét, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{m+5}{2}\\x_1x_2=\dfrac{c}{a}=\dfrac{-3m^2+10m-3}{2}\end{matrix}\right.\)
Ta có: \(x_1^2+x_2^2-\left(x_1+x_2\right)+x_1x_2=4\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2-\left(x_1+x_2\right)+x_1x_2=4\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-x_1x_2-\left(x_1+x_2\right)=4\)
\(\Leftrightarrow\left(\dfrac{m+5}{2}\right)^2-\dfrac{-3m^2+10m-3}{2}-\dfrac{m+5}{2}=4\)
\(\Leftrightarrow\dfrac{m^2+10m+25}{4}-\dfrac{-6m+20m-6}{4}-\dfrac{2m+10}{4}=\dfrac{16}{4}\)
\(\Leftrightarrow m^2+10m+25+6m^2-20m+6-2m-10=16\)
\(\Leftrightarrow7m^2-12m+5=0\)
Ta có: \(a+b+c=7+\left(-12\right)+5=0\)
Phương trình có hai nghiệm:
\(m_1=1;m_2=\dfrac{c}{a}=\dfrac{5}{7}\)
Vậy \(m_1=1;m_2=\dfrac{c}{a}=\dfrac{5}{7}\) thì phương trình đã cho có hai nghiệm thỏa mãn: \(x_1^2+x_2^2-\left(x_1+x_2\right)+x_1x_2=4\)
