\(a,m=-1\Rightarrow pt=\dfrac{1}{2}x^2+x+\dfrac{1}{2}-4-1=0\Leftrightarrow\dfrac{1}{2}x^2+x-\dfrac{9}{2}=0\)
Có : \(\Delta=b^2-4ac=1-4.\dfrac{1}{2}.\left(-\dfrac{9}{2}\right)=10>0\Rightarrow\) Pt có 2 nghiệm pb \(x_1,x_2\)
\(\left\{{}\begin{matrix}x_1=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-1-\sqrt{10}}{1}=-1-\sqrt{10}\\x_2=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-1+\sqrt{10}}{1}=-1+\sqrt{10}\end{matrix}\right.\)
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\(b,\) Theo Vi-ét, ta có :
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2\\x_1x_2=\dfrac{c}{a}=\dfrac{\dfrac{1}{2}m^2+4m-1}{\dfrac{1}{2}}=m^2+8m-2\end{matrix}\right.\)
Có : \(\dfrac{1}{x_1}+\dfrac{1}{x_2}=x_1+x_2\Leftrightarrow\dfrac{x_1+x_2}{x_1x_2}=x_1+x_2\)
\(\Leftrightarrow\dfrac{2}{m^2+8m-2}=2\Leftrightarrow\dfrac{2-2\left(m^2+8m-2\right)}{m^2+8m-2}=0\Leftrightarrow2-2m^2-16m+4=0\\ \Leftrightarrow-2m^2-16m+6=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}m_1=-4+\sqrt{19}\\m_2=-4-\sqrt{19}\end{matrix}\right.\)
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