已知x2+xy=4,xy+y2=12,求代数式x2-y2与x2+2xy+y2的值各为多少

题目

已知x2+xy=4,xy+y2=12,求代数式x2-y2与x2+2xy+y2的值各为多少
格式正确

答案

X2+xy-(xy+y2)=4-12
x2+xy-xy-y2=-8
x2-y2=-8
x2+xy+xy+y2=4+12
x2+2xy+y2=16