若x^2-x=4,y^2-y=4,且x不等于y.则x^2+2xy+y^2=?

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若x^2-x=4,y^2-y=4,且x不等于y.则x^2+2xy+y^2=?

若x^2-x=4,y^2-y=4,且x不等于y.则x^2+2xy+y^2=?
若x^2-x=4,y^2-y=4,且x不等于y.则x^2+2xy+y^2=?

若x^2-x=4,y^2-y=4,且x不等于y.则x^2+2xy+y^2=?

x^2-x=4,y^2-y=4
则x,y都是方程 A²-A-4=0的根.
利用韦达定理
x+y=1
∴ x^2+2xy+y^2=(x+y)²=1
如果没有学过韦达定理
x^2-x=4, ①
y^2-y=4 ②
①-②
x²-y²-(x-y)=0
∴(x-y)(x+y)=x-y
∵ x≠y
∴ x-y≠0
∴ x+y=1
∴ x^2+2xy+y^2=(x+y)²=1

x^2-x=4,y^2-y=4
∴x,y是方程z²-z-4=0的两根
∴x+y=1
xy=-4
∴x^2+2xy+y^2=(x+y)²=1

x^2-x=1
y^2-y=1
两式相减得
x^2-y^2-x+y=0
(x-y)(x+y)-(x-y)=0
(x-y)(x+y-1)=0
因为x≠y
所以x+y-1=0 x+y=1
x^2+2xy+y^2
=(x+y)^2
=1^2
=1