题目内容
有这样一道计算题:“计算(2x3-3x2y-2xy2)-(x3-2xy2+y3)+(-x3+3x2y-y3)的值,其中x=
,y=-1”,甲同学把x=
看错成x=-
,但计算结果仍正确,亲爱的同学,你能解释是怎么一回事吗?
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(2x3-3x2y-2xy2)-(x3-2xy2+y3)+(-x3+3x2y-y3)
=2x3-3x2y-2xy2-x3+2xy2-y3-x3+3x2y-y3=(2-1-1)x3+(-3+3)x2y+(-2+2)xy2+(-1-1)y3
=-2y3,
故代数式的值与x的取值无关.
=2x3-3x2y-2xy2-x3+2xy2-y3-x3+3x2y-y3=(2-1-1)x3+(-3+3)x2y+(-2+2)xy2+(-1-1)y3
=-2y3,
故代数式的值与x的取值无关.
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