已知:x+y=1,求(x2-y2)2-2(x2+y2)的值.
题型:解答题难度:一般来源:不详
已知:x+y=1,求(x2-y2)2-2(x2+y2)的值. |
答案
解法一:原式=(x+y)2(x-y)2-2(x2+y2), ∵x+y=1, ∴原式=(x-y)2-2(x2+y2) =x2-2xy+y2-2x2-y2 =-x2-2xy-y2 =-(x+y)2 =-1; 解法二:∵x+y=1, ∴y=1-x, ∴原式=[x2-(1-x)2]2-2[x2+(1-x)2] =(2x-1)2-2(2x2-2x+1) =4x2-4x+1-4x2+4x-2 =-1. |
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