Question

若x+y=-1,则x^4+5x^3y+x^2y+8x^2y^2+xy^2+5xy^3+y^4=( ）

Answer 1

=(x^4+x^3y)+(y^4+xy^3)+(4x^3y+4x^2y^2)+(4xy^3+4x^2y^2)+x^2y+xy^2 =x^3(x+y)+y^3(x+y)+4x^2y(y+x)+4xy^2(y+x)+x^2y+xy^2 =-x^3-y^3-4x^2y-4xy^2+xy^2+x^2y =-x^3-y^3-3x^2y-3xy^2 =-(x^3+x^2y)-(y^3+xy^2)-2x^2y-2xy^2 =-x^2(x+y)-y^2(x+y)-2xy(x+y) =x^2+y^2+2xy =(x+y)^2 =1 记得采纳我的答案哦，祝你学习进步