"已知x=3,y=-1.求代数式(x³+3x²y-5xy²+6x³+1)-(2x³-y³-2xy²-x²y-2)-(4x²y+7x³+y³-4xy²-1)的值时,误把“x=3,y=-1”写成“x=3.y=1”,但其计算结果仍然是正确的,
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/01 06:05:16
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"已知x=3,y=-1.求代数式(x³+3x²y-5xy²+6x³+1)-(2x³-y³-2xy²-x²y-2)-(4x²y+7x³+y³-4xy²-1)的值时,误把“x=3,y=-1”写成“x=3.y=1”,但其计算结果仍然是正确的,
"已知x=3,y=-1.求代数式
(x³+3x²y-5xy²+6x³+1)-(2x³-y³-2xy²-x²y-2)-(4x²y+7x³+y³-4xy²-1)的值时,误把“x=3,y=-1”写成“x=3.y=1”,但其计算结果仍然是正确的,请解释其中的原因.
"已知x=3,y=-1.求代数式(x³+3x²y-5xy²+6x³+1)-(2x³-y³-2xy²-x²y-2)-(4x²y+7x³+y³-4xy²-1)的值时,误把“x=3,y=-1”写成“x=3.y=1”,但其计算结果仍然是正确的,
去括号,合并同类项并按x的降幂排列
(x³+3x²y-5xy²+6x³+1)-(2x³-y³-2xy²-x²y-2)-(4x²y+7x³+y³-4xy²-1)
=-2x³+xy²+4
可见,只含有y的偶数次方项,y=-1与y=1的结果都是一样的