已知x2+xy+y=0,y2+xy+x=0,求x+y的值那几个2均为次数
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已知x2+xy+y=0,y2+xy+x=0,求x+y的值那几个2均为次数
已知x2+xy+y=0,y2+xy+x=0,求x+y的值
那几个2均为次数
已知x2+xy+y=0,y2+xy+x=0,求x+y的值那几个2均为次数
x²+xy+y=0 (1)
y²+xy+x=0 (2)
(1)-(2)
x²-y²+y-x=0
(x+y)(x-y)-(x-y)=0
(x-y)(x+y-1)=0
x-y=0或x+y-1=0 或x-y=0且x+y-1=0
x-y=0时,x=y,代入(1),得3y²=0 解得y=0 此时x+y=0+0=0
x+y-1=0时,x+y=1(特例:x=y=1/2时,x-y=0、x+y-1=0同时成立)
综上,得x+y=0或x+y=1
x2+xy+y-(y2+xy+x)=x^2-y^2+y-x=(x-y)(x+y)-(x-y)=(x-y)(x+y-1)=0
所以,x-y=0或x+y-1=0
即x+y=1