a)Nếu m=0 thì pt\(\Rightarrow-x-2=0\Rightarrow x=-2\)
\(\Rightarrow\)Pt có nghiệm duy nhất
\(\Rightarrow m=0\left(loại\right)\)
Nếu \(m\ne0\) thì pt có hai nghiệm
\(\Leftrightarrow\Delta\ge0\Rightarrow\left(2m+1\right)^2-4\cdot m\cdot\left(m-2\right)\ge0\)
\(\Rightarrow4m^2+4m+1-4m^2+8m\ge0\)
\(\Rightarrow m\ge-\dfrac{1}{12}\) thì pt có hai nghiệm \(x_1,x_2\)
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2m+1}{m}\\x_1x_2=\dfrac{m-2}{m}\end{matrix}\right.\)
Ta có \(2\left(x_1+x_2\right)+x_1x_2=4x_1x_2\left(x_1+x_2\right)\)
Thay vào ta được \(\dfrac{4m+2}{m}+\dfrac{m-2}{m}=\dfrac{4\left(m-2\right)\left(2m+1\right)}{m^2}\)
\(\Rightarrow4m^2+2m+m^2-2m=4\left(2m^2-3m-2\right)\)
\(\Leftrightarrow-3m^2+12m+8=0\Leftrightarrow x=\dfrac{6\pm2\sqrt{15}}{3}\)(tm)
b)Hệ thức Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{2m+1}{m}\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{m-2}{m}\end{matrix}\right.\left(m\ne\right)0\) (1)
\(2x_1+2x_2+x_1\cdot x_2=4x_1^2x_2+4x_1x_2^2\)
\(\Rightarrow2\left(x_1+x_2\right)+x_1\cdot x_2=4x_1x_2\left(x_1+x_2\right)\)(2)
Thay (1) vào (2) ta đc:
\(2\cdot\dfrac{2m+1}{m}+\dfrac{m-2}{m}=4\cdot\dfrac{m-2}{m}\cdot\left(\dfrac{2m+1}{m}\right)\)
\(\Rightarrow4m+2+m-2=\left(4m-8\right)\left(2m+1\right)\)
\(\Rightarrow8m^2-17m-8=0\)
\(\Rightarrow m=\dfrac{17\pm\sqrt{545}}{16}\)
