$\rm \text{Thay x=1 vào A,x=-1 vào B thì:}\\1+2.m.1+m^2=1+(2m+1).m+m^2\\\to m^2+2m+1=2m^2+m^2+m+1\\\to 3m^2+m+1=m^2+2m+1\\\to 2m^2-m=0\\\to m(2m-1)=0\\\to \left[ \begin{array}{l}x=0\\x=\dfrac{1}{2}\end{array} \right.\\\text{Vậy x=0 hoặc x=1/2 thì A(1)=B(-1)}$
Em tham khảo:
Ta có
$A(1)=B(-1)$
⇔$1^{2}+2m+m^2=1+(2m+1).m+m^2$
⇔$1+2m+m^2=1+$$2m^{2}+m+m^2$
⇔$2m^{2}-m=0$
⇔$m.(2m-1)=0$
⇔\(\left[ \begin{array}{l}m=0\\m=\dfrac{1}{2}\end{array} \right.\)
Học tốt
Đáp án:
$\rm \text{Thay x=1 vào A,x=-1 vào B thì:}\\1+2.m.1+m^2=1+(2m+1).m+m^2\\\to m^2+2m+1=2m^2+m^2+m+1\\\to 3m^2+m+1=m^2+2m+1\\\to 2m^2-m=0\\\to m(2m-1)=0\\\to \left[ \begin{array}{l}x=0\\x=\dfrac{1}{2}\end{array} \right.\\\text{Vậy x=0 hoặc x=1/2 thì A(1)=B(-1)}$