\(\Delta=\left(2m+1\right)^2-4m\left(m-2\right)=12m+1\)
a/ Với \(m=0\) pt có nghiệm
Với \(m\ne0\) để pt có nghiệm \(\Rightarrow12m+1\ge0\Rightarrow m\ge-\frac{1}{12}\)
Khi đó, theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=\frac{2m+1}{m}\\x_1x_2=\frac{m-2}{m}\end{matrix}\right.\)
b/ \(\left(x_1+x_2\right)^2-2x_1x_2=22\)
\(\Leftrightarrow\left(2+\frac{1}{m}\right)^2-2\left(1-\frac{2}{m}\right)=22\)
\(\Leftrightarrow\left(\frac{1}{m}\right)^2+8\left(\frac{1}{m}\right)-20=0\Rightarrow\left[{}\begin{matrix}\frac{1}{m}=2\\\frac{1}{m}=-10\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}m=\frac{1}{2}\\m=-\frac{1}{10}\left(l\right)\end{matrix}\right.\)
c/ \(x_1^2-x_2^2=13\)
\(\Leftrightarrow x_1-x_2=\frac{13}{x_1+x_2}=\frac{13m}{2m+1}\)
Kết hợp Viet ta được: \(\left\{{}\begin{matrix}x_1+x_2=\frac{2m+1}{m}\\x_1-x_2=\frac{13m}{2m+1}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\frac{17m^2+4m+1}{2m\left(2m+1\right)}\\x_2=\frac{-9m^2+4m+1}{2m\left(2m+1\right)}\end{matrix}\right.\)
\(x_1x_2=\frac{m-2}{m}\Rightarrow\frac{\left(17m^2+4m+1\right)\left(-9m^2+4m+1\right)}{4m\left(2m+1\right)^2}=m-2\)
\(\Leftrightarrow169m^4-48m^3-52m^2-16m-1=0\)
Pt này cho nghiệm rất xấu, chắc đề bài sao đó
