Sửa đề: \(x_1^2x_2+x_1x_2^2=24\)
\(\Delta=\left[-\left(m+5\right)\right]^2-4\left(-m+6\right)=m^2+10m+25+4m-24=m^2+14m+1\)
PT có 2 nghiệm pb \(\Leftrightarrow\Delta>0\Leftrightarrow m^2+14m+1>0\)\(\Leftrightarrow\left[{}\begin{matrix}m< -7-4\sqrt{3}\\m>-7+4\sqrt{3}\end{matrix}\right.\)
Theo Vi-ét ta có : \(\left\{{}\begin{matrix}x_1+x_2=m+5\\x_1x_2=-m+6\end{matrix}\right.\)
Theo đề bài có: \(x_1^2x_2+x_1x_2^2=24\)
\(\Leftrightarrow x_1x_2\left(x_1+x_2\right)=24\)
\(\Leftrightarrow\left(-m+6\right)\left(m+5\right)=24\)
\(\Leftrightarrow-m^2+m+30=24\)
\(\Leftrightarrow m^2-m-6=0\)
\(\Leftrightarrow\left(m-3\right)\left(m+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m=3\\m=-2\left(L\right)\end{matrix}\right.\)
Vậy m = 3 là GT cần tìm
