若x^2+xy+y=5,y^2+xy+x=7,则x+y的值为球解答TVT

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若x^2+xy+y=5,y^2+xy+x=7,则x+y的值为球解答TVT
若x^2+xy+y=5,y^2+xy+x=7,则x+y的值为
球解答TVT

若x^2+xy+y=5,y^2+xy+x=7,则x+y的值为球解答TVT
解
x²+xy+y=5
y²+xy+x=7
两式相加得：
x²+xy+y+y²+xy+x=5+7
即x²+2xy+y²+y+x=12
即(x+y)²+(x+y)-12=0
∴[(x+y)+4][(x+y)-3]=0
∴(x+y)+4=0或(x+y)-3=0
∴x+y=-4或x+y=3

x^2+xy+y=5, -----------------------(1)
y^2+xy+x=7, -----------------------(2)
(1)+(2)得：
x^2+xy+y+y^2+xy+x=5+7
(x+y)^2+(x+y)=12
(x+y)^2+(x+y)-12=0
((x+y)+4)((x+y)-3)=0
(x+y)=-4 或 (x+y)=3

相加答案是3或-4

x^2+xy+y=5 (1)
y^2+xy+x=7 (2)
(1)+(2):
x^2 + y^2+2xy+x+y=12
(x+y)^2 + (x+y) =12
设x+y = t
t^2 + t - 12 = 0
(t-3)(t+4)=0
t = 3 或 t = -4
x+y = 3或-4

二式相加得X²＋Y²＋2XY＋X＋Y=12∴（X＋Y）²＋X＋Y＝12∴X＋Y＝3或-4 望采纳

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