四道初一计算题1、(y^2+2)^2-(y+2)(y-2)(y^2+4)2、(x+y)(y-x)-(2+y^2)(1-x
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/11/12 00:18:08
四道初一计算题
1、(y^2+2)^2-(y+2)(y-2)(y^2+4)
2、(x+y)(y-x)-(2+y^2)(1-x)
3、(m+2)(m-2)(m^2-2m+4)(m^2+2m+4)
4、(x+y)(x-y)+(y-z)(y+z)+(z-x)(x+z)
PS:要过程~~!
1、(y^2+2)^2-(y+2)(y-2)(y^2+4)
2、(x+y)(y-x)-(2+y^2)(1-x)
3、(m+2)(m-2)(m^2-2m+4)(m^2+2m+4)
4、(x+y)(x-y)+(y-z)(y+z)+(z-x)(x+z)
PS:要过程~~!
1、(y^2+2)^2-(y+2)(y-2)(y^2+4)
=(y^2+2)^2-(y^2-4)(y^2+4)
=y^4+4-y^4+16
=4y^2+20
2、(x+y)(y-x)-(2+y^2)(1-x)
=y^2-x^2-2+2x-y^2+xy^2
=xy^2-x^2+2x-2
3、(m+2)(m-2)(m^2-2m+4)(m^2+2m+4)
=(m^2-4)(8m^2+16)
=8(m^4-2m^2-8)
=8(m^2-4)(m^2+2)
=8(m^2+2)(m+2)(m-2)
4、(x+y)(x-y)+(y-z)(y+z)+(z-x)(x+z)
=(x^2-y^2)+(y^2-z^2)+(z^2-x^2)
=0
=(y^2+2)^2-(y^2-4)(y^2+4)
=y^4+4-y^4+16
=4y^2+20
2、(x+y)(y-x)-(2+y^2)(1-x)
=y^2-x^2-2+2x-y^2+xy^2
=xy^2-x^2+2x-2
3、(m+2)(m-2)(m^2-2m+4)(m^2+2m+4)
=(m^2-4)(8m^2+16)
=8(m^4-2m^2-8)
=8(m^2-4)(m^2+2)
=8(m^2+2)(m+2)(m-2)
4、(x+y)(x-y)+(y-z)(y+z)+(z-x)(x+z)
=(x^2-y^2)+(y^2-z^2)+(z^2-x^2)
=0
计算题(2x+y)(x+2y)-(2x+3y)(3x-2y)
[(-x-y)(-x+y)-(x+y)^2-x(y-y^2)}÷1/2y
计算题:(9x-4y)(-3x+2y)(-2y-3x)
1、x(x-y)(x+y)-x(x+y)^2
计算题:(2x+3y)^2-(4x-9y)(4x+9y)+(2x-3y)^2
(1)(x^2/x)-y-x-y
计算题:(-2x+3y)^2
1) (2x+5y)(2x-5y)(-4x^2-25y^2) 2) [(x+y)(x-y)-(x-y)^2+2y(x-y
初一数学计算题(2x+3y)(2x-3y)
1 4(x-y+1)+y(y-2x)
把4(x-y+1)+y(y-2x)因式分解
4(x-y+1)+y(y-2x)