已知：x+y=1,xy=-1/2,求：(x+y)²(x-y)+2y²(x+y)的值.
问题描述：
已知：x+y=1,xy=-1/2,求：(x+y)²(x-y)+2y²(x+y)的值.
答
因为：x+y=1
所以：(x+y)^2=1
即：x^2+y^2+2xy=1
因为xy=-1/2
所以：x^2+y^2=2
所以：(x+y)^2(x-y)+2y^2(x+y)
=(x+y)(x^2-y^2+2y^2)
=(x+y)(x^2+y^2)
=1*2=2
所以所求代数式的值是2
答
等于2
答
(x+y)²(x-y)+2y²(x+y)=（x+y）[(x+y)(x-y)+2y²]=(x+y)(x²-y²+2y²)=(x+y)(x²+y²)=x²+y²=（x+y）²-2xy=1-2（-1/2）=1+1=2
答
(x+y)²(x-y)+2y²(x+y)
=(x+y)(x²-y²+2y²)
=(x+y)(x²+y²)
=(x+y)[(x+y)²-2xy]
=1×【1-2×(-1/2)】
=2