a/ Bạn tự giải
b/ \(\Delta=\left(2m+3\right)^2+4\left(2m+4\right)>0\)
\(\Leftrightarrow4m^2+20m+25>0\Leftrightarrow\left(2m+5\right)^2>0\)
\(\Rightarrow m\ne-\frac{5}{2}\)
Khi đó theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m+3\\x_1x_2=-2m-4\end{matrix}\right.\)
\(\left|x_1\right|+\left|x_2\right|=5\)
\(\Leftrightarrow x_1^2+x_2^2+2\left|x_1x_2\right|=25\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|=25\)
\(\Leftrightarrow\left(2m+3\right)^2+2\left(2m+4\right)+2\left|2m+4\right|-25=0\)
\(\Leftrightarrow m^2+4m-2+\left|m+2\right|=0\)
\(\Leftrightarrow\left(m+2\right)^2+\left|m+2\right|-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|m+2\right|=-3\left(l\right)\\\left|m+2\right|=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}m=0\\m=-4\end{matrix}\right.\)
