题目内容
口算:
(1)1+2+4+8+16+32+64+128+256+512=
(2)1000001×999999=
(3)(1+2+3+…+2007+2008+2007+…+3+2+1)÷2008=
(4)(2005+2003+…+3+1)-(2+4+6+…+2002+2004)=
(1)1+2+4+8+16+32+64+128+256+512=
(2)1000001×999999=
(3)(1+2+3+…+2007+2008+2007+…+3+2+1)÷2008=
(4)(2005+2003+…+3+1)-(2+4+6+…+2002+2004)=
分析:(1)前边数乘2就是后边的数,可以设原式加上1再减去1,然后再进一步计算即可;
(2)可以把1000001=(1000000+1),再根据乘法分配律进行计算即可;
(3)原式是有两个1+2+3+…+2007相加再加上2008的和除以2008,可以根据等差数列先求出1+2+3+…+2007的和,再进一步计算即可;
(4)2005+2003+…+3+1有1003个数,2+4+6+…+2002+2004有1002个数,把括号去掉即2005+2003+…+3+1-2-4-6-…-2002-2004,然后相邻的两个数相减即(2005-2004)+(2003-2002)+…+(7-6)+(5-4)+(3-2)+1,就可以得到1003个1,就是1003.
(2)可以把1000001=(1000000+1),再根据乘法分配律进行计算即可;
(3)原式是有两个1+2+3+…+2007相加再加上2008的和除以2008,可以根据等差数列先求出1+2+3+…+2007的和,再进一步计算即可;
(4)2005+2003+…+3+1有1003个数,2+4+6+…+2002+2004有1002个数,把括号去掉即2005+2003+…+3+1-2-4-6-…-2002-2004,然后相邻的两个数相减即(2005-2004)+(2003-2002)+…+(7-6)+(5-4)+(3-2)+1,就可以得到1003个1,就是1003.
解答:解:(1)1+2+4+8+16+32+64+128+256+512,
=(1+2+4+8+16+32+64+128+256+512)+1-1,
=(2+2+4+8+16+32+64+128+256+512)-1,
=(4+4+8+16+32+64+128+256+512)-1,
=(8+8+16+32+64+128+256+512)-1,
=(16+16+32+64+128+256+512)-1,
=(32+32+64+128+256+512)-1,
=(64+64+128+256+512)-1,
=(128+128+256+512)-1,
=(256+256+512)-1,
=(512+512)-1,
=1023;
(2)1000001×999999,
=(1000000+1)×999999,
=1000000×999999+1×999999,
=999999000000+999999;
=999999999999;
(3)(1+2+3+…+2007+2008+2007+…+3+2+1)÷2008,
=[(1+2007)×2007÷2×2+2008]÷2008,
=[2008×2007+2008]÷2008,
=2008×(2007+1)÷2008,
=2008×2008÷2008,
=2008;
(4)(2005+2003+…+3+1)-(2+4+6+…+2002+2004),
=2005+2003+…+3+1-2-4-6-…-2002-2004,
=(2005-2004)+(2003-2002)+…+(7-6)+(5-4)+(3-2)+1,
=1+1+…+1+1+1+1,
=1003.
=(1+2+4+8+16+32+64+128+256+512)+1-1,
=(2+2+4+8+16+32+64+128+256+512)-1,
=(4+4+8+16+32+64+128+256+512)-1,
=(8+8+16+32+64+128+256+512)-1,
=(16+16+32+64+128+256+512)-1,
=(32+32+64+128+256+512)-1,
=(64+64+128+256+512)-1,
=(128+128+256+512)-1,
=(256+256+512)-1,
=(512+512)-1,
=1023;
(2)1000001×999999,
=(1000000+1)×999999,
=1000000×999999+1×999999,
=999999000000+999999;
=999999999999;
(3)(1+2+3+…+2007+2008+2007+…+3+2+1)÷2008,
=[(1+2007)×2007÷2×2+2008]÷2008,
=[2008×2007+2008]÷2008,
=2008×(2007+1)÷2008,
=2008×2008÷2008,
=2008;
(4)(2005+2003+…+3+1)-(2+4+6+…+2002+2004),
=2005+2003+…+3+1-2-4-6-…-2002-2004,
=(2005-2004)+(2003-2002)+…+(7-6)+(5-4)+(3-2)+1,
=1+1+…+1+1+1+1,
=1003.
点评:根据给出的数据,选择合适的方法是计算这样题目的关键,然后再进一步计算即可.
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