\(pt:x^2-2\left(m-1\right)x-m-3=0\\ Thay\cdot m=-3:pt\Leftrightarrow x^2+8x=0\Leftrightarrow x\left(x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)
Để pt có 2 nghiệm thì:
\(\Delta>0\Leftrightarrow\left(-2m+2\right)^2-4.\left(-m-3\right)=4-8m+4m^2+4m+12=4m^2-4m+16=\left(2m-1\right)^2+15>0\forall m\)
Theo hệ thức Vi-et:
\(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\\x_1x_2=-m-3\end{matrix}\right.\)
\(x_1^2+x_2^2=10\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=10\Leftrightarrow\left(2m-2\right)^2-2\left(-m-3\right)=10\Leftrightarrow4m^2-8m+4+2m+6=10\Leftrightarrow4m^2-6m=0\Leftrightarrow m\left(4m-6\right)=0\Leftrightarrow\left[{}\begin{matrix}m=0\\m=\frac{3}{2}\end{matrix}\right.\)