试说明:无论x,y取何值时,代数式(x3+3x2y-5+6y3)+(y3+2xy2+x2y-2x3)-(4x2y-x3-
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试说明:无论x,y取何值时,代数式(x3+3x2y-5+6y3)+(y3+2xy2+x2y-2x3)-(4x2y-x3-3xy2+7y2)的值是常数.
要具体过程,注意格式
要具体过程,注意格式
![试说明:无论x,y取何值时,代数式(x3+3x2y-5+6y3)+(y3+2xy2+x2y-2x3)-(4x2y-x3-](/uploads/image/z/18806574-30-4.jpg?t=%E8%AF%95%E8%AF%B4%E6%98%8E%3A%E6%97%A0%E8%AE%BAx%2Cy%E5%8F%96%E4%BD%95%E5%80%BC%E6%97%B6%2C%E4%BB%A3%E6%95%B0%E5%BC%8F%28x3%2B3x2y-5%2B6y3%29%2B%28y3%2B2xy2%2Bx2y-2x3%29-%284x2y-x3-)
(x^3+3x^2y-5xy^2+6y^3)+(y^3+2xy^2+x^2y-2x^3)-(4x^2y-x^3-3xy^2+7y^3)
=x^3+3x^2y-5xy+6y^3+y^3+2xy^2+x^2y-2x^3-4x^2y+x^3+3xy^2-7y^3
=(x^3-2x^3+x^3)+(3x^2y+x^2y-4x^2y)+(-5xy^2+2xy^2+3xy^2)+(6y^3+y^3-7y^3)
=0+0+0=0
可知:代数式的值与x,y的取值无关.
=x^3+3x^2y-5xy+6y^3+y^3+2xy^2+x^2y-2x^3-4x^2y+x^3+3xy^2-7y^3
=(x^3-2x^3+x^3)+(3x^2y+x^2y-4x^2y)+(-5xy^2+2xy^2+3xy^2)+(6y^3+y^3-7y^3)
=0+0+0=0
可知:代数式的值与x,y的取值无关.
已知A=x3+3x2y-5xy2+6y3-1,B=y3+2xy2+x2y-2x3+2,C=x3-4x2y+3xy2-7y
有这样一道题:“计算(2x3-3x2y-2xy2)-(x3-2xy2+y3)+(-x3+3x2y-y3)的值,其中x=1
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因式分解:x3-y3-x2y+xy2
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