题目内容
计算:
(1)20122-2×2010×2012+20102
(2)(-
)-1+(π-3)0-(-1)2
(3)(a+b)(a-b)(a2+b2)(a4+b4)
(4)(a-2b-3)(a-2b+3)
(1)20122-2×2010×2012+20102
(2)(-
| 1 |
| 2012 |
(3)(a+b)(a-b)(a2+b2)(a4+b4)
(4)(a-2b-3)(a-2b+3)
(1)原式=2=22=4;
(2)原式=-2012+1-1=-2012;
(3)原式=(a2-b2)(a2+b2)(a4+b4)=(a4-b4)(a4+b4)=a8-b8;
(4)原式=(a-2b)2-9=a2-4ab+4b2-9.
(2)原式=-2012+1-1=-2012;
(3)原式=(a2-b2)(a2+b2)(a4+b4)=(a4-b4)(a4+b4)=a8-b8;
(4)原式=(a-2b)2-9=a2-4ab+4b2-9.
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