已知|x+2y+1|+(2x-y-2)平方=0,求(2x-y)平方-2(2x-y)(x+2y)+(x+2y)平方的值
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已知|x+2y+1|+(2x-y-2)平方=0,求(2x-y)平方-2(2x-y)(x+2y)+(x+2y)平方的值
已知|x+2y+1|+(2x-y-2)平方=0,求(2x-y)平方-2(2x-y)(x+2y)+(x+2y)平方的值
已知|x+2y+1|+(2x-y-2)平方=0,求(2x-y)平方-2(2x-y)(x+2y)+(x+2y)平方的值
因为|x+2y+1|>=0,(2x-y-2)²>=0
所以要使式子的值等于0,必须有
x+2y+1=0,2x-y-2=0
所以x+2y=-1,2x-y=2
(2x-y)²-2(2x-y)(x+2y)+(x+2y)²
=[(2x-y)-(x+2y)]²
=(2-(-1))²
=3²
=9
绝对值和平方大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立
所以两个都等于0
所以x+2y+1=0 (1)
2x-y-2=0 (2)
所以x+2y=-1
2x-y=2
(2x-y)^2-2(2x-y)(x+2y)+(x+2y)^2
=[(2x-y)-(x+2y)]^2
=[2-(-1)]^2
=3^2
=9
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