a.\(m=2\)
\(\Leftrightarrow x^2-2\left(2-1\right)x-2.2=0\)
\(\Leftrightarrow x^2-2x-4=0\)
\(\Delta=\left(-2\right)^2-4.\left(-4\right)\)
\(=4+16=20>0\)
--> pt có 2 nghiệm
\(\left\{{}\begin{matrix}x_1=\dfrac{2+\sqrt{20}}{2}=1+\sqrt{5}\\x_2=\dfrac{2-\sqrt{20}}{2}=1-\sqrt{5}\end{matrix}\right.\)
b.\(\Delta=\left(-2\left(m-1\right)\right)^2-4\left(-2m\right)=4m^2-8m+4+8m=4m^2+4>0\)
--> pt luôn có 2 nghiệm phân biệt
Theo hệ thức Vi-ét, ta có:\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)=2m-2\\x_1x_2=-2m\end{matrix}\right.\) (1)
\(x_1^2+x_1-x_2=5-2m\)
\(\Leftrightarrow x_1^2+x_1-x_2=3-\left(2m-2\right)\)
\(\Leftrightarrow x_1^2+x_1-x_2=3-x_1-x_2\)
\(\Leftrightarrow x_1^2+2x_1-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\) ( Vi-ét )
Thế x1=1 vào(1) \(\rightarrow\left\{{}\begin{matrix}1+x_2=2m-2\\1x_2=-2m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_2-2m=-3\\x_2+2m=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_2=-\dfrac{3}{2}\\m=\dfrac{3}{4}\end{matrix}\right.\)
Thế x1=-3 vào (1) \(\rightarrow\left\{{}\begin{matrix}-3+x_2=2m-2\\-3x_2=-2m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_2-2m=1\\-3x_2+2m=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_2=-\dfrac{1}{2}\\m=-\dfrac{3}{4}\end{matrix}\right.\)
Vậy \(m=\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)
