分式的约分与通分问题约分 (x²+xy)/(x+y)² x²-y²/(x&sup

问题描述:

分式的约分与通分问题
约分 (x²+xy)/(x+y)²
x²-y²/(x²-y²)
通分2xy/(x+y)²与x/x²-y²
x-y/2x+2y与xy/(x+y)²
1个回答 分类:数学 2014-10-02

问题解答:

我来补答
(x²+xy)/(x+y)² 分子提取公因式x
=x(x+y)/(x+y)² 分子分母同时除以(x+y)
=x/(x+y)
x²-y²/(x²-y²)好像题目不对
2xy/(x+y)²与x/x²-y² 公因式为(x+y)² (x-y)
2xy/(x+y)²=2xy(x-y)/{(x+y)²(x-y)}
x/(x²-y²)=x(x+y)/{(x+y)²(x-y)} 说明x²-y² =(x+y)(x-y)
x-y/2x+2y与xy/(x+y)² 公因式为2(x+y)²
x-y/2x+2y
=(x-y)/{2(x+y)}
=(x+y)(x-y) /{2(x+y))(x+y)}
=x+y)(x-y) /{2(x+y)²}
xy/(x+y)²
=2xy/{2(x+y)²}
 
 
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