P/t có : \(\Delta'=\left(m-2\right)^2-\left(m^2+2m-3\right)=-6m+7\)
Để p/t có 2 no x1 ; x2 p/b thì : \(\Delta'>0\Leftrightarrow-6m+7>0\Leftrightarrow m< \dfrac{7}{6}\) (3)
Theo Viet ta có : \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-2\right)\\x_1x_2=m^2+2m-3\end{matrix}\right.\)
Ta có : \(\dfrac{1}{x_1}+\dfrac{1}{x_2}=\dfrac{x_1+x_2}{5}\)
\(\Leftrightarrow\left(x_1+x_2\right)\left(\dfrac{1}{x_1x_2}-\dfrac{1}{5}\right)=0\) \(\Leftrightarrow2\left(m-2\right)\left(m^2+2m-3-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m-2=0\\m^2+2m-8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}m=2\\\left(m+4\right)\left(m-2\right)=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}m=2\\m=-4\end{matrix}\right.\) (1)
Mặt khác : \(x_1;x_2\ne0\Rightarrow x_1x_2\ne0\) \(\Leftrightarrow m^2+2m-3\ne0\) \(\Leftrightarrow\left(m+3\right)\left(m-1\right)\ne0\)
\(\Leftrightarrow m\ne-3;1\) (2)
Từ (1) ; (2) ; (3) suy ra : m = -4
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