Lời giải:
Ta có: \(A(x)=x^2-(3m+3)x+m^2\)
\(\Rightarrow A(-1)=1+(3m+3)+m^2=m^2+3m+4\)
\(B(x)=x^3+(5m-7)x+m^2\)
\(\Rightarrow B(2)=8+2(5m-7)+m^2=m^2+10m-6\)
Do đó để \(A(-1)=B(2)\Leftrightarrow m^2+3m+4=m^2+10m-6\)
\(\Leftrightarrow 3m+4=10m-6\Leftrightarrow 10=7m\Leftrightarrow m=\frac{10}{7}\)