a/ Bạn tự giải
b/ Trừ vế cho vế \(\Leftrightarrow\left\{{}\begin{matrix}y=m\\x+y=2m+1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=m\\x=m+1\end{matrix}\right.\)
\(x^2+y^2< m^2+6m+6\)
\(\Leftrightarrow m^2+\left(m+1\right)^2< m^2+6m+6\)
\(\Leftrightarrow m^2-4m-5< 0\)
\(\Leftrightarrow\left(m+1\right)\left(m-5\right)< 0\Rightarrow-1< m< 5\)
a) \(m=-3\) ta có hệ phương trình:
\(\left\{{}\begin{matrix}x+2y=-8\\x+y=-5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=-3\\x=-2\end{matrix}\right.\)
Vậy khi \(m=-3\) thì hệ có nghiệm \(\left(x;y\right)=\left(-2;-3\right)\)
b)
\(\left\{{}\begin{matrix}x+2y=3m+1\\x+y=2m+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=m\\x=m+1\end{matrix}\right.\)
Ta có: \(x^2+y^2< m^2+6m+6\)
\(\Rightarrow m^2+2m+1+m^2< m^2+6m+6\)
\(\Leftrightarrow m^2-4m-5< 0\)
\(\Leftrightarrow\left(m+1\right)\left(m-5\right)< 0\)
\(\Leftrightarrow-1< m< 5\)
Vậy \(-1< m< 5\)
