a) \(x^2-\left(2m+1\right)x+m^2+m-2=0\) (1)
Thay m = 2 vào phương trình (1), ta được:
\(x^2-\left(2\cdot2+1\right)x+2^2+2-2=0\)
\(\Leftrightarrow x^2-5x+4=0\)
\(\Leftrightarrow x^2-x-4x+4=0\)
\(\Leftrightarrow x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
b) \(\Delta=\left(2m+1\right)^2-4\left(m^2+m-2\right)=9>0;\forall m\)
\(\Rightarrow\) Phương trình (1) có hai nghiệm phân biệt với mọi m
Theo hệ thức Vi-ét: \(\left\{{}\begin{matrix}x_1+x_2=2m+1\\x_1x_2=m^2+m-2\end{matrix}\right.\)
Ta có: \(x_1\left(x_1-2x_2\right)+x_2\left(x_2-3x_1\right)=9\)
\(\Leftrightarrow\left(x_1^2+x_2^2\right)-2x_1x_2-3x_1x_2=9\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-7x_1x_2=9\)
\(\Rightarrow\left(2m+1\right)^2-7\left(m^2+m-2\right)=9\)
\(\Leftrightarrow4m^2+4m+1-7m^2-7m+14=9\)
\(\Leftrightarrow-3m^2-3m+6=0\)
\(\Leftrightarrow m^2+m-2=0\Leftrightarrow\left[{}\begin{matrix}m=1\\m=-2\end{matrix}\right.\)
$\text{#}Toru$
