\(\Delta'=\left(m+1\right)^2-\left(m^2-2\right)\)
\(\Leftrightarrow\Delta'=m^2+2m+1-m^2+2\)
\(\Leftrightarrow\Delta'=2m+3\)
a)Để pt có 2 No phân biệt trái dấu
\(\Leftrightarrow\left\{{}\begin{matrix}a.c< 0\\\Delta'>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}1.\left(m^2-2\right)< 0\\2m+3>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{2}< m< \sqrt{2}\\m>\frac{-3}{2}\end{matrix}\right.\)
\(\Leftrightarrow-\sqrt{2}< m< \sqrt{2}\)
b) để phương trình có 2 No phân biệt
\(\Leftrightarrow\Delta'>0
\)
\(\Leftrightarrow m>\frac{-3}{2}\)
Theo định lí Vi-et
\(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\left(1\right)\\x_1.x_2=m^2-2\left(2\right)\end{matrix}\right.\)
Và \(x_1^2+x_2^2+x_1.x_2=2\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-x_1.x_2=2\)(3)
Thay (1)và (2) vào (3)
\(\left(3\right)\Rightarrow\left(2m+2\right)^2-m^2+2=2\)
\(\Leftrightarrow m=\left\{{}\begin{matrix}-2\left(l\right)\\0\left(tm\right)\end{matrix}\right.\)