当x=2^1001.y=(-2)^1001时,代数式4x^2-8xy+4y^2的值为?
问题描述：
当x=2^1001.y=(-2)^1001时,代数式4x^2-8xy+4y^2的值为?
答
4x^2-8xy+4y^2=(2x-2y)²=2²（x-y）²
x-y=2x=2^1002
2²（x-y）²=2^2006
答
x=2^1001.y=(-2)^1001
4x^2-8xy+4y^2
=4(x-y)^2
=4*(2*2^1001)^2
=4^1003
答
4x^2-8xy+4y^2
=4(x^2-2xy+y^2)
=4(x-y)^2
=4(2^1001+2^1001)^2
=2^2*(2^1002)^2
=2^2006