a) Khi \(m=-1\) thì ta có pt:
\(x^2+x-2=0\)
\(\Delta=1^2-4\cdot1\cdot\left(-2\right)=9>0\)
\(x_1=\dfrac{-1+\sqrt{9}}{2}=1\)
\(x_2=\dfrac{-1-\sqrt{9}}{2}=-2\)
b) Để pt có 2 nghiệm phân biệt thì \(\Delta>0\)
\(\Leftrightarrow\left(-m\right)^2-4\cdot1\cdot\left(-2\right)=m^2+8>0\forall m\)
Theo vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=-2\end{matrix}\right.\)
Ta có: \(x_1-2x_2=5\)
\(\Leftrightarrow x_1+x_2=5+3x_2\)
\(\Leftrightarrow m-5=3x_2\)
\(\Leftrightarrow x_2=\dfrac{m-5}{3}\)
\(\Rightarrow x_1=-2:\dfrac{m-5}{3}=\dfrac{-6}{m-5}\)
\(x_1-2x_2=5\)
\(\Rightarrow\dfrac{-6}{m-5}-2\cdot\dfrac{m-5}{3}=5\)
\(\Leftrightarrow\dfrac{-6}{m-5}-\dfrac{2\left(m-5\right)}{3}=5\)
\(\Leftrightarrow\dfrac{-18-2\left(m-5\right)^2}{3\left(m-5\right)}=5\)
\(\Leftrightarrow-18-2\left(m^2-10m+25\right)=15\left(m-5\right)\)
\(\Leftrightarrow-18-2m^2+20m-50=15m-75\)
\(\Leftrightarrow-2m^2+5m+7=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{7}{2}\\m=-1\end{matrix}\right.\)
