\(x^2+2x+m-3=0\left(x+1\right)^2=4-m\Leftrightarrow\)(1)
(1) có 2 No.có nghiệm: \(\Leftrightarrow4-m>0\Rightarrow m< 4\) (*)
Hai nghiệm là : \(\left[\begin{matrix}x_1=1-\sqrt{4-m}\\x_2=1+\sqrt{4-m}\end{matrix}\right.\Rightarrow\left(I\right)\left\{\begin{matrix}x_1x_2=m-3\\x_1+x_2=2\end{matrix}\right.\)
\(x^3_1x_2+x_1x^3_2=x_1x_2\left(x^2_1+x_2^2\right)=x_1x_2\left[\left(x_1+x_2\right)^2-2x_1x_2\right]=-6\)
Thay x1&x2 theo m từ (I) vào ta có :
\(\left(m-3\right)\left[2^2-2\left(m-3\right)\right]=4\left(m-3\right)-2\left(m-3\right)^2=-6\)
\(\Leftrightarrow\left(m-3\right)^2-2\left(m-3\right)-6=0\Rightarrow\left\{\begin{matrix}m_1=4-\sqrt{7}\\m_2=4+\sqrt{7}\end{matrix}\right.\)
từ (*) \(\Rightarrow m=4+\sqrt{7}\)
