a: A(-1)=0
=>-m+2=0
=>m=2
b: B(0)=6 nên 0xm+n=6
=>n=6
=>B(x)=mx+6
B(2)=0
=>2m+6=0
hay m=-3
c: C(0)=2 nên p=2
Vậy: \(C\left(x\right)=mx^2+nx+2\)
Theo đề, ta có hệ:
\(\left\{{}\begin{matrix}m+n+2=0\\4m+2n+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m+n=-2\\4m+2n=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m=1\\n=-3\end{matrix}\right.\)



