a) Ta có △\(=b^2-4ac=\left[2\left(m+1\right)\right]^2-4.1.\left(-m^2\right)=4\left(m+1\right)^2+4m^2\ge0\Rightarrow\)phương trình luôn có nghiệm \(x_1,x_2\)
b) Theo định lí Vi-ét ta có
\(\left\{{}\begin{matrix}x_1+x_2=\frac{-b}{a}=\frac{-2m-2}{1}=-2m-2\\x_1x_2=\frac{c}{a}=\frac{-m^2}{1}=-m^2\end{matrix}\right.\)
Ta lại có \(x^2_1+x_2^2=4\Leftrightarrow x^2_1+2x_1x_2+x_2^2-2x_1x_2=4\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=4\Leftrightarrow\left[-\left(2m+2\right)\right]^2-2\left(-m^2\right)=4\Leftrightarrow4m^2+8m+4+2m^2=4\Leftrightarrow6m^2+8m=0\Leftrightarrow3m^2+4m=0\Leftrightarrow m\left(3m+4\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}m=0\\m=-\frac{4}{3}\end{matrix}\right.\)
Vậy m=0 hoặc m=\(\frac{-4}{3}\) thì \(x_1^{^2}+x_2^2=4\)
