a/ Pt có 2 nghiệm
\(→Δ=[-(2m-1)]^2-4.1.(m^2-2m)≥0\\↔4m^2-4m+1-4m^2+8m≥0\\↔4m+1≥0\\↔m≥-\dfrac{1}{4}\)
b/ Theo Vi-ét:
\(\begin{cases}x_1+x_2=2m-1\\x_1x_2=m^2-2m\end{cases}\)
\(2x_1+2x_2+1+x_1x_2+2x_1^2x_2+2x_1x_2^2\\=2(x_1+x_2)+1+x_1x_2+2x_1x_2(x_1+x_2)\\=(x_1+x_2)(2+2x_1x_2)+(1+x_1x_2)\\=2(x_1+x_2)(1+x_1x_2)+(1+x_1x_2)\\=[2(x_1+x_2)+1)(x_1x_2+1)\\=[2(2m-1)+1](m^2-2m+1)\\=(4m-2+1)(m-1)^2\\=(4m-1)(m-1)^2\\→(4m-1)(m-1)^2=0\\↔4m-1=0\,\, or\,\, m-1=0\\↔m=\dfrac{1}{4}(tm)\,\, or\,\, m=1(tm)\)
Vậy $m=\bigg\{\dfrac{1}{4};1\bigg\}$
