x+xy+y=1,x^2+x^2*y^2+x^2=17求x,y的值答案是x=y=(3±√17)÷2
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x+xy+y=1,x^2+x^2*y^2+x^2=17求x,y的值答案是x=y=(3±√17)÷2x+xy+y=1,x^2+x^2*y^2+x^2=17求x,y的值答案是x=y=(3±√17)÷2x+
x+xy+y=1,x^2+x^2*y^2+x^2=17求x,y的值答案是x=y=(3±√17)÷2
x+xy+y=1,x^2+x^2*y^2+x^2=17
求x,y的值
答案是x=y=(3±√17)÷2
x+xy+y=1,x^2+x^2*y^2+x^2=17求x,y的值答案是x=y=(3±√17)÷2
因为x+xy+y=1平方后把x^2+x^2y^2+y^2=17代入
可得xy(x+y+1)=-8
设x+y=a,xy=b,则:
a+b=1
b(a+1)=-8
解得a=3,b=-2或a=-3,b=4
再解出x=(3+√17)/2,y=(3-√17)/2
或x=(3-√17)/2,y=(3+√17)/2
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