计算(1)(-2x^2)^5/(-x^3)^3+(2x)^4(2)【5xy^2(x^2-3xy)-(-3x^2y^2)^
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计算(1)(-2x^2)^5/(-x^3)^3+(2x)^4(2)【5xy^2(x^2-3xy)-(-3x^2y^2)^3]/5(xy)^2
如题
如题
(-2x^2)^5/(-x^3)^3+(2x)^4
=(-32x^10)/(-x^9)
=32x
[5xy^2(x^2-3xy)-(-3x^2y^2)^3]/5(xy)^2
=(5x^3y^2-15x^2y^3+27x^6y^6)/(5x^2y^2)
=[5x^2y^2(x-3y+27/5x^4y^4)]/(5x^2y^2)
=x-3y+27/5x^4y^4
再问: 不好意思,第一题打错了,应该是(-2x^2)^5/(-x^3)^2+(2x)4
再答: (-2x^2)^5/(-x^3)^2+(2x)^4 =(-32x^10)/(x^6)+8x^4 =-32x^4+8x^4 =-24x^4
再问: (2x)^4怎么就等于8x^4.....再怎么说也应该是16x^4吧...
再答: 额,搞错了 (-2x^2)^5/(-x^3)^2+(2x)^4 =(-32x^10)/(x^6)+16x^4 =-32x^4+16x^4 =-16x^4
=(-32x^10)/(-x^9)
=32x
[5xy^2(x^2-3xy)-(-3x^2y^2)^3]/5(xy)^2
=(5x^3y^2-15x^2y^3+27x^6y^6)/(5x^2y^2)
=[5x^2y^2(x-3y+27/5x^4y^4)]/(5x^2y^2)
=x-3y+27/5x^4y^4
再问: 不好意思,第一题打错了,应该是(-2x^2)^5/(-x^3)^2+(2x)4
再答: (-2x^2)^5/(-x^3)^2+(2x)^4 =(-32x^10)/(x^6)+8x^4 =-32x^4+8x^4 =-24x^4
再问: (2x)^4怎么就等于8x^4.....再怎么说也应该是16x^4吧...
再答: 额,搞错了 (-2x^2)^5/(-x^3)^2+(2x)^4 =(-32x^10)/(x^6)+16x^4 =-32x^4+16x^4 =-16x^4
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