三道巧算①[a+(b-c)][a-(b-c)]②(a²+1)(2^4+1)(2^8+1)…(2^32+1)③2009*2
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①[a+(b-c)][a-(b-c)]=(a+b-c)(a-b+c)=a²-(b-c)²={再自己展开喽————
②第一项 应是(2²+1)吧
(2²+1)(2^4+1)(2^8+1)…(2^32+1)
=1x (2²+1)(2^4+1)(2^8+1)…(2^32+1)
=(2-1) (2²+1)(2^4+1)(2^8+1)…(2^32+1)
=1/3(2+1)(2-1)(2²+1)(2^4+1)(2^8+1)…(2^32+1)
=1/3(2²-1)(2²+1)(2^4+1)(2^8+1)…(2^32+1)
=1/3(2^4-1)(2^4+1)(2^8+1)…(2^32+1)
=1/3(2^8-1)(2^8+1）…(2^32+1)
=1/3（2^64-1）
③2009*2010*2011-2010³
=2010（2010-1)(2010+1)-2010³
=2010(2010²-1)-2010³
=2010³-1-2010³
=1
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