题目内容
若x(x4+y4)=y5,x2(x+y)≠y3,则x3+y3=
1
1
.分析:根据x(x4+y4)=y5,进行变形经过因式分解即可求出答案.
解答:解:x5-y5
=x5-x4y+x4y-xy4+xy4-y5
=x5-y5-xy(x3+y3)+xy(x3+y3)
∴-xy(x3+y3)+xy(x3+y3)=0
x3+y3=1.
故答案为:1.
=x5-x4y+x4y-xy4+xy4-y5
=x5-y5-xy(x3+y3)+xy(x3+y3)
∴-xy(x3+y3)+xy(x3+y3)=0
x3+y3=1.
故答案为:1.
点评:本题主要考查因式分解的应用,解答本题的关键是根据题干条件进行变形从而得到结果.
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