若一个长方形的周长为28,两边周长分别为x、y,且满足x³+x²y-xy²-y³=0,试求这个长方

问题描述:

若一个长方形的周长为28,两边周长分别为x、y,且满足x³+x²y-xy²-y³=0,试求这个长方

x2(x+y)-y2(x+y)=0
(x+y)(x2-y2)=0
(x+y)(x+y)(x-y)=0
长方形的边为正数,所以x+y不等于0
所以 x-y=0 x=y
所以实际是正方形,边长7 面积49

x³+x²y-xy²-y³=0
x²(x+y)-y²(x+y)=(x+y)(x²-y²)=(x+y)(x+y)(x-y)=0
x+y=28/2=14
x-y=0
x=y=7

x³+x²y-xy²-y³=x²(x+y)-y²(x+y)=(x²-y²)(x+y)=(x-y)(x+y)²=0
x+y>0.则x-y=0
x=y=28/4=7