Question

Cho 2 đa thức 

P(x)=\(x^2+2mx+m^2\)

Q(x)=\(x^2+\left(2m+1\right)x+m^2\)

Tìm m biết P(1)=Q(-1)

Answer 1

\(P\left(1\right)=Q\left(-1\right)\)

\(\Leftrightarrow1^2+2m+m^2=\left(-1\right)^2+\left(2m+1\right)\cdot\left(-1\right)+m^2\)

\(\Leftrightarrow m^2+2m+1=m^2-2m-1+1\)

=>2m+1=-2m

=>4m=-1

hay m=-1/4