\(\Delta'=\left(m+1\right)^2-m^2+3m=5m+1>0\Rightarrow m>-\frac{1}{5}\)
Theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2-3m\end{matrix}\right.\)
a/ \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\2x_1-3x_2=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x_1+3x_2=6\left(m+1\right)\\2x_1-3x_2=8\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=\frac{6m+14}{5}\\x_2=\frac{4m-4}{5}\end{matrix}\right.\)
\(x_1x_2=m^2-3m\)
\(\Leftrightarrow\left(\frac{6m+14}{5}\right)\left(\frac{4m-4}{5}\right)=m^2-3m\)
Bạn tự khai triển và giải pt bậc 2 này
b/ \(\left|x_1-x_2\right|=4\)
\(\Leftrightarrow\left(x_1-x_2\right)^2=16\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=16\)
\(\Leftrightarrow4\left(m+1\right)^2-4\left(m^2-3m\right)=16\)
\(\Leftrightarrow5m+1=4\)
c/
\(\left|x_1\right|+\left|x_2\right|=3\)
\(\Leftrightarrow\left(\left|x_1\right|+\left|x_2\right|\right)^2=9\)
\(\Leftrightarrow x_1^2+x_2^2+2\left|x_1x_2\right|=9\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+2\left|x_1x_2\right|=9\)
\(\Leftrightarrow4\left(m+1\right)^2-2\left(m^2-3m\right)+2\left|m^2-3m\right|=9\)
- Với \(m^2-3m\ge0\Leftrightarrow\left[{}\begin{matrix}m\ge3\\-\frac{1}{5}< m\le0\end{matrix}\right.\)
\(\Rightarrow4\left(m+1\right)^2-2\left(m^2-3m\right)+2\left(m^2-3m\right)=9\)
\(\Leftrightarrow4\left(m+1\right)^2=9\Rightarrow\left(m+1\right)^2=\frac{9}{4}\)
\(\Rightarrow\left[{}\begin{matrix}m+1=\frac{3}{2}\\m+1=-\frac{3}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=\frac{1}{3}\left(l\right)\\m=-\frac{5}{2}< -\frac{1}{5}\left(l\right)\end{matrix}\right.\)
- Với \(m^2-3m< 0\Rightarrow0< m< 3\)
\(\Rightarrow4\left(m+1\right)^2-2\left(m^2-3m\right)-2\left(m^2-3m\right)=9\)
\(\Leftrightarrow20m-5=0\Rightarrow m=\frac{1}{4}\) (thỏa mãn)
