①[(x+y)^2-(x+y)(x-y)]/2y②(-1/3a^3b^2)^2(-2ab^2)^3/(1/9a^4b^4)^2高手帮帮忙
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①[(x+y)^2-(x+y)(x-y)]/2y②(-1/3a^3b^2)^2(-2ab^2)^3/(1/9a^4b^4)^2高手帮帮忙
①[(x+y)^2-(x+y)(x-y)]/2y②(-1/3a^3b^2)^2(-2ab^2)^3/(1/9a^4b^4)^2高手帮帮忙
①[(x+y)^2-(x+y)(x-y)]/2y②(-1/3a^3b^2)^2(-2ab^2)^3/(1/9a^4b^4)^2高手帮帮忙
①[(x+y)^2-(x+y)(x-y)]/2y
=[(x+y)(x+y-x+y)]/2y
=[(x+y)*2y]/2y
=x+y
②(-1/3a^3b^2)^2(-2ab^2)^3/(1/9a^4b^4)^2
=(1/9a^6b^4)(-8a^3b^6)/(1/81a^8b^8)
=(-8/9a^9b^10) /(1/81a^8b^8)
=-72ab^2
我告诉你解题思路,你自己计算,这两道题并不难,不能只想着求助,要自己思考。
第一题,把分子上的因式都打开,然后合并同类项,就可以约分了。
第二题,分子分母全都是相乘的关系,所以也是一样,把括号都打开,约分就可以了
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化简:[(x+y)(x-y)(x+y)^2-2y(x-2y)](-2y) 谢谢!