已知x²+y²+5=2x+4y,求代数式(2x²-(x+y)(x-y))x((x+y-1)(x-y+1)+(1-2y))
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已知x²+y²+5=2x+4y,求代数式(2x²-(x+y)(x-y))x((x+y-1)(x-y+1)+(1-2y))
已知x²+y²+5=2x+4y,求代数式(2x²-(x+y)(x-y))x((x+y-1)(x-y+1)+(1-2y))
已知x²+y²+5=2x+4y,求代数式(2x²-(x+y)(x-y))x((x+y-1)(x-y+1)+(1-2y))
已知x²+y²+5=2x+4y
所以(x-1)²+(y-2)²=0
故x=1,y=2
所以(2x²-(x+y)(x-y))×((x+y-1)(x-y+1)+(1-2y))
=(2-(1+2)(1-2))×((1+2-1)(1-2+1)+(1-4))
=(2+3)×(-3)
=-15
X=1,Y=2代入得—15