a, Thay m=2 vào (1) ta có:
\(x^2-\left(2.2-1\right)x+2.2-2=0\\
\Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\
\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(b,\Delta=\left[-\left(2m-1\right)\right]^2-4\left(2m-2\right)\\ =4m^2-4m+1-8m+8\\ =4m^2-12m+9\\ =\left(2m-3\right)^2\ge0\)
Suy ra pt luôn có 2 nghiệm
Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2m-1\\x_1x_2=2m-2\end{matrix}\right.\)
\(\left(2x_1+x_2\right)\left(2x_2+x_1\right)=2\\ \Leftrightarrow4x_1x_2+2x_2^2+2x_1^2+x_1x_2=2\\ \Leftrightarrow2\left(x_1^2+x_2^2\right)+5x_1x_2=2\\ \Leftrightarrow2\left(x_1+x_2\right)^2-4x_1x_2+5x_1x_2=2\\ \Leftrightarrow2.\left(2m-1\right)^2+x_1x_2=2\\ \Leftrightarrow2\left(4m^2-4m+1\right)+2m-2-2=0\\ \Leftrightarrow8m^2-8m+2+2m-4=0\\ \Leftrightarrow8m^2-6m-2=0\\ \Leftrightarrow\left[{}\begin{matrix}m=1\\m=-\dfrac{1}{4}\end{matrix}\right.\)
